{"id":29411,"date":"2022-03-02T16:36:46","date_gmt":"2022-03-02T11:06:46","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=29411"},"modified":"2022-03-02T16:36:46","modified_gmt":"2022-03-02T11:06:46","slug":"ap-board-9th-class-maths-solutions-chapter-2-ex-2-5-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-2-ex-2-5-in-telugu\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 2 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15 \u0c35\u0c3f\u0c2d\u0c1c\u0c28 Ex 2.5"},"content":{"rendered":"

SCERT AP 9th Class Maths Solutions<\/a> Chapter 2 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15 \u0c35\u0c3f\u0c2d\u0c1c\u0c28 Ex 2.5 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 2nd Lesson \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15 \u0c35\u0c3f\u0c2d\u0c1c\u0c28 Exercise 2.5<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c281.
\n\u0c24\u0c17\u0c3f\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c32\u0c2c\u0c4d\u0c26\u0c3e\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\ni) (x + 5) (x + 2)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(x + 5) (x + 2) = x2<\/sup> + (5 + 2)x + 5 \u00d7 2
\n[\u2235 (x + a) (x + b) = x2<\/sup> + (a + b)x + ab]
\n= x2<\/sup> + 7x + 10<\/p>\n

ii) (x – 5) (x – 5)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(x – 5) (x – 5) = (x – 5)2<\/sup>
\n= x2<\/sup> – 2(x) (5) + 52<\/sup>
\n[\u2235 (x – y)2<\/sup> = x2<\/sup> – 2xy + y2<\/sup>]
\n= x2<\/sup> – 10x + 25<\/p>\n

iii) (3x + 2) (3x – 2)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(3x + 2) (3x – 2) = (3x)2<\/sup> – (2)2<\/sup>
\n[\u2235 (x + y) (x – y) = x2<\/sup> – y2<\/sup>]
\n= 9x2<\/sup> – 4<\/p>\n

iv) \\(\\left(x^{2}+\\frac{1}{x^{2}}\\right)\\left(x^{2}-\\frac{1}{x^{2}}\\right)\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(\\left(x^{2}+\\frac{1}{x^{2}}\\right)\\left(x^{2}-\\frac{1}{x^{2}}\\right)\\)
\n= (x2)2 – \\(\\left(\\frac{1}{x^{2}}\\right)^{2}\\)
\n[\u2235 (x + y) (x – y) = x2<\/sup> – y2<\/sup>]
\n= x4<\/sup> – \\(\\frac{1}{x^{4}}\\)<\/p>\n

v) (1 + x) (1 + y)
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1 + x) (1 + x)
\n= (1 + x)2<\/sup> = 12<\/sup> + 2 (1) (x) + x2<\/sup>
\n[\u2235 (x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup>]
\n= 1 + 2x + x2<\/sup><\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c282.
\n\u0c17\u0c41\u0c23\u0c15\u0c3e\u0c30\u0c02 \u0c1a\u0c47\u0c2f\u0c15\u0c41\u0c02\u0c21\u0c3e\u0c28\u0c47 \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c32\u0c2c\u0c4d\u0c26\u0c3e\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) 101 \u00d7 99
\n\u0c38\u0c3e\u0c27\u0c28.
\n101 \u00d7 99 = (100 + 1) (100 – 1)
\n= 1002<\/sup> – 12<\/sup> = 10000 – 1 = 9999<\/p>\n

ii) 999 \u00d7 999
\n\u0c38\u0c3e\u0c27\u0c28.
\n999 \u00d7 999 = 9992<\/sup>
\n= (1000 – 1)2<\/sup>
\n= 10002 – 2 \u00d7 (1000) \u00d7 1 + 12<\/sup>
\n= 1000000 – 2000 + 1
\n= 998001<\/p>\n

iii) 50\\(\\frac {1}{2}\\) \u00d7 49\\(\\frac {1}{2}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n50\\(\\frac {1}{2}\\) \u00d7 49\\(\\frac {1}{2}\\) = \\(\\left(50+\\frac{1}{2}\\right)\\left(50-\\frac{1}{2}\\right)\\)
\n= 502<\/sup> – \\(\\left(\\frac{1}{2}\\right)^{2}\\)
\n= 2500 – \\(\\frac {1}{4}\\)
\n= 2499\\(\\frac {3}{4}\\)
\n= \\(\\frac {9999}{4}\\)<\/p>\n

iv) 501 \u00d7 501
\n\u0c38\u0c3e\u0c27\u0c28.
\n501 \u00d7 501 = (500 + 1) (500 + 1)
\n= (500 + 1)2<\/sup>
\n= 5002 + 2 \u00d7 (500) \u00d7 1 + 12<\/sup>
\n= 250000 + 1000 + 1
\n= 251001<\/p>\n

v) 30.5 \u00d7 29.5
\n\u0c38\u0c3e\u0c27\u0c28.
\n30.5 \u00d7 29.5 = (30 + 0.5) (30 – 0.5)
\n= 302<\/sup> – (0.5)2<\/sup>
\n= 900 – 0.25 = 899.75<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c283.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c28\u0c41 \u0c24\u0c17\u0c3f\u0c28 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41\u0c32\u0c28\u0c41 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c3f\u0c2d\u0c1c\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) 16x2<\/sup> + 24xy + 9y2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n16x2<\/sup> + 24xy + 9y2<\/sup>
\n= (4x)2<\/sup> + 2 (4x) (3y) + (3y)2<\/sup>
\n= (4x + 3y)2<\/sup> = (4x + 3y) (4x + 3y)
\n[\u2235 (x + y)2<\/sup> \u2261 x2<\/sup> + 2xy + y2<\/sup>]<\/p>\n

ii) 4y2<\/sup> – 4y + 1
\n\u0c38\u0c3e\u0c27\u0c28.
\n4y2<\/sup> – 4y + 1 = (2y)2<\/sup> – 2 (2y) (1) + (1)2<\/sup>
\n= (2y – 1)2<\/sup>
\n= (2y – 1) (2y – 1) [\u2235 (x + y)2<\/sup> \u2261 x2<\/sup> – 2xy + y2<\/sup>]<\/p>\n

iii) 4x2<\/sup> + \\(\\frac{y^{2}}{25}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n4x2<\/sup> + \\(\\frac{y^{2}}{25}\\) = (2x)2<\/sup> – \\(\\left(\\frac{y}{5}\\right)^{2}\\)
\n= \\(\\left(2 x+\\frac{y}{5}\\right)\\left(2 x-\\frac{y}{5}\\right)\\)
\n[\u2235 (x + y) (x – y) \u2261 x2<\/sup> – y2<\/sup>]<\/p>\n

iv) 18a2<\/sup> – 50
\n\u0c38\u0c3e\u0c27\u0c28.
\n18a2<\/sup> – 50 = 2(9a2<\/sup> – 25)
\n= 2[(3a)2<\/sup> – (5)2<\/sup>]
\n= 2 (3a + 5) (3a – 5)
\n[\u2235 x2<\/sup> – y2<\/sup> \u2261 (x + y) (x – y)]<\/p>\n

v) x2<\/sup> + 5x + 6
\n\u0c38\u0c3e\u0c27\u0c28.
\nx2<\/sup> + 5x + 6 = x2<\/sup> + (3 + 2)x + 3 \u00d7 2
\n= (x + 3) (x + 2)
\n[\u2235 (x + a) (x + b) \u2261 x2<\/sup> + (a + b) x + a . b]<\/p>\n

vi) 3p2<\/sup> – 24p + 36
\n\u0c38\u0c3e\u0c27\u0c28.
\n3p2<\/sup> – 24p + 36
\n= 3[p2<\/sup> – 8p + 12]
\n= 3[p2<\/sup> + (-6)p + (-2)p + (-6) (-2)]
\n= 3 [p (p – 6) – 2 (p – 6)]
\n= 3 (p – 6) (p – 2)
\n[\u2235 (x + a) (x + b) \u2261 x2<\/sup> + (a + b) x + ab]<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c284.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c24\u0c17\u0c3f\u0c28 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c35\u0c3f\u0c38\u0c4d\u0c24\u0c30\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) (x + 2y + 4z)2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(x + 2y + 4z)2<\/sup>
\n= (x)2<\/sup> + (2y)2 + (4z)2<\/sup> + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
\n= x2<\/sup> + 4y2<\/sup> + 16z2<\/sup> + 4xy + 16yz + 8zx
\n[\u2235 (x + y + z)2<\/sup> \u2261 x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx]<\/p>\n

ii) (2a – 3b)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(2a – 3b)3<\/sup>
\n= (2a)3<\/sup> -3 (2a)2<\/sup> (3b) + 3(2a)(3b)2<\/sup> – (3b)3<\/sup>
\n= 8a3<\/sup> – 3(4a2<\/sup>) (3b) + 3 (2a) (9b2<\/sup>) – 27b3<\/sup>
\n= 8a3<\/sup> – 36a2<\/sup>b + 54ab2<\/sup> – 27b3<\/sup>
\n[\u2235 (a – b)3<\/sup> \u2261 a3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup> – b3<\/sup>]
\n(\u0c32\u0c47\u0c26\u0c3e)
\n(2a – 3b)3<\/sup>
\n= (2a)3<\/sup> – (3b)3<\/sup> – 3(2a) (3b) (2a – 3b)
\n= 8a3<\/sup> – 27b3<\/sup> – 18ab (2a – 3b)
\n[\u2235 (a – b)3<\/sup> \u2261 a3<\/sup> – b3<\/sup> – 3ab (a – b)]<\/p>\n

iii) (-2a + 5b – 3c)2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(- 2a + 5b – 3c)2<\/sup>
\n= (-2a)2<\/sup> + (5b)2<\/sup> + (-3c)2<\/sup> + 2 (-2a) (5b) + 2 (5b) (-3c) + 2 (-3c) (-2a)
\n= 4a2<\/sup> + 25b2<\/sup> + 9c2<\/sup> – 20ab – 30bc + 12ca
\n[\u2235 (x + y + z)2<\/sup> \u2261 x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx]<\/p>\n

iv) \\(\\left(\\frac{a}{4}-\\frac{b}{2}+1\\right)^{2}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP<\/p>\n

v) (p + 1)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(p + 1)3<\/sup>
\n= (p)3<\/sup> + 3 (p)2<\/sup> (1) + 3 (p) (1)2<\/sup> + (1)3<\/sup>
\n[\u2235 (x + y)3<\/sup> \u2261 x3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> + y3<\/sup>]
\n= p3<\/sup> + 3p2<\/sup> + 3p +1<\/p>\n

vi) \\(\\left(x-\\frac{2}{3} y\\right)^{3}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\"AP
\n[\u2235 (x – y)3<\/sup> \u2261 x3<\/sup> – 3x2<\/sup>y + 3xy2<\/sup> – y3]
\n= x3<\/sup> – 2x2<\/sup>y + \\(\\frac{4 x y^{2}}{3}-\\frac{8}{27} y^{3}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c285.
\n\u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c3f\u0c2d\u0c1c\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) 25x2<\/sup> + 16y2<\/sup> + 4z2<\/sup> – 40xy + 16yz – 20xz
\n\u0c38\u0c3e\u0c27\u0c28.
\n25x2<\/sup> + 16y2<\/sup> + 4z2<\/sup> – 40xy + 16yz – 20xz
\n= (5x)2<\/sup> + (-4y)2<\/sup> + (-2z)2<\/sup> + 2(5x) (-4y) + 2 (-4y) (-2z) + 2 (-2z) (5x)
\n[\u2235 x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx = (x + y + z)2<\/sup>]
\n= (5x – 4y – 2z)2<\/sup> = (-5x + 4y + 2z)2<\/sup><\/p>\n

ii) 9a2<\/sup> + 4b2<\/sup> + 16c2<\/sup> + 12ab – 16bc – 24ca
\n\u0c38\u0c3e\u0c27\u0c28.
\n9a2<\/sup> + 4b2<\/sup> + 16c2<\/sup> + 12ab – 16bc – 24ca
\n= (3a)2<\/sup> + (2b)2<\/sup> + (-4c)2<\/sup> + 2 (3a) (2b) + 2 (2b) (-4c) + 2(-4c) (3a)
\n= (3a + 2b – 4c)2<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c286.
\na + b + c = 9 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 ab + bc + ca = 26 \u0c05\u0c2f\u0c3f\u0c28 a2<\/sup> + b2<\/sup> + c2<\/sup> \u0c35\u0c3f\u0c32\u0c41\u0c35 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 a + b + c = 9
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32\u0c3e \u0c35\u0c30\u0c4d\u0c17\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e,
\n(a + b + c)2<\/sup> = 92<\/sup>
\n\u21d2 a2<\/sup> + b2<\/sup> + c2 <\/sup>+ 2 (ab + bc + ca) = 81
\n\u21d2 a2<\/sup> + b2<\/sup> + c2<\/sup> = 81 – 2 (ab + bc + ca) (\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c2e\u0c41)
\n= 81 – 2 \u00d7 26
\n= 81 – 52
\n= 29<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c287.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) (99)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(99)3<\/sup> = (100 – 1)3<\/sup>
\n= 1003<\/sup> – 3 (100)2<\/sup> (1) + 3 (100) (1)2<\/sup> – 13<\/sup>
\n[\u2235 (x – y)3<\/sup> \u2261 x3<\/sup> – 3x2<\/sup>y + 3xy2<\/sup> – y3<\/sup>]
\n= 1000000 – 30000 + 300 – 1
\n= 970299<\/p>\n

ii) (102)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(102)3<\/sup> = (100 + 2)3<\/sup>
\n= 1003<\/sup> + 3 (100)2<\/sup> (2) + 3 (100) (2)2<\/sup> + 23<\/sup>
\n[\u2235 (x + y)3<\/sup> \u2261 x3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> + y]
\n= 1000000 + 60000 + 1200 + 8
\n= 1061208<\/p>\n

iii) (998)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(998)3<\/sup> = (1000 – 2)3<\/sup>
\n[\u2235 (x – y)3<\/sup> \u2261 x3<\/sup> – 3x2<\/sup>y + 3xy2<\/sup> – y3<\/sup>]
\n= 10003<\/sup> – 3(1000)2<\/sup> (2) + 3(1000) (2)2<\/sup> – 23<\/sup>
\n= 1000000000 – 6000000 + 12000 – 8
\n= 994011992<\/p>\n

iv) (1001)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(1001)3<\/sup> = (1000 + 1)3<\/sup>
\n[\u2235 (x + y)3<\/sup> = x3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> + y3<\/sup>]
\n= 10003<\/sup> + 3(1000)2<\/sup> (1) + 3(1000) (1)2<\/sup> + 13<\/sup>
\n= 1000000000 + 3000000 + 3000 + 1
\n= 1003003001<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c288.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c28\u0c3f\u0c28\u0c3f \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c3f\u0c2d\u0c1c\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41\u0c28\u0c41 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup> \u0c17\u0c3e \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c17\u0c3e
\n= (2a)3<\/sup> + (b)3<\/sup> + 3 (2a)2<\/sup> (b) + 3 (2a) (b)2<\/sup>
\n= (2a + b)3<\/sup><\/p>\n

ii) 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41\u0c28\u0c41 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup> \u0c17\u0c3e \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c17\u0c3e
\n= (2a)3<\/sup> – (b)3<\/sup> – 3 (2a)2<\/sup> (b) + 3 (2a) (b)2<\/sup>
\n= (2a – b)3<\/sup><\/p>\n

iii) 1 – 64a3<\/sup> – 12a + 48a2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n1 – 64a3<\/sup> – 12a + 48a2<\/sup>
\n= (1)3<\/sup> – (4a)3<\/sup> – 3(1)2<\/sup> (4a) + 3(1) (4a)2<\/sup>
\n= (1 – 4a)3<\/sup><\/p>\n

iv) \\(8 \\mathrm{p}^{3}-\\frac{12}{5} \\mathrm{p}^{2}+\\frac{6}{25} \\mathrm{p}-\\frac{1}{125}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\\(8 \\mathrm{p}^{3}-\\frac{12}{5} \\mathrm{p}^{2}+\\frac{6}{25} \\mathrm{p}-\\frac{1}{125}\\)
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c289.
\ni) x3<\/sup> + y3<\/sup> = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\nii) x3<\/sup> – y3<\/sup> = (x – y) (x2<\/sup> + xy + y2<\/sup>) \u0c32\u0c28\u0c41 \u0c38\u0c30\u0c3f\u0c1a\u0c42\u0c21\u0c02\u0c21\u0c3f. \u0c05\u0c26\u0c47 \u0c35\u0c3f\u0c27\u0c02\u0c17\u0c3e \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c30\u0c02 \u0c1a\u0c47\u0c38\u0c3f \u0c32\u0c2c\u0c4d\u0c26\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c2a\u0c30\u0c3f\u0c36\u0c40\u0c32\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f. \u0c35\u0c40\u0c1f\u0c3f\u0c28\u0c3f \u0c15\u0c42\u0c21\u0c3e \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41\u0c32\u0c28\u0c3f \u0c2d\u0c3e\u0c35\u0c3f\u0c02\u0c1a\u0c35\u0c1a\u0c4d\u0c1a\u0c41\u0c28\u0c3e ?
\n1) x3<\/sup> + y3<\/sup> = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 x3<\/sup> + y3<\/sup> = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\nL.H.S = x3<\/sup> + y3<\/sup>
\nR.H.S = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n= x (x2<\/sup> – xy + y2<\/sup>) + y (x2<\/sup> – xy + y2<\/sup>)
\n= x3<\/sup> – x2<\/sup>y + xy2<\/sup> + x2<\/sup>y – xy2<\/sup> + y3<\/sup>
\n= x3<\/sup> + y3<\/sup> = L.H.S
\n\u2234 L.H.S = R.H.S
\nx = 3, y = 2 \u0c17\u0c3e \u0c24\u0c40\u0c38\u0c41\u0c15\u0c4a\u0c28\u0c3f\u0c28
\nL.H.S = 33<\/sup> + 23<\/sup> = 27 + 8 = 35
\nR.H.S= (3 + 2) (32<\/sup> – 3 \u00d7 2 + 22<\/sup>)
\n= 5 \u00d7 (9 – 6 + 4) = 5 \u00d7 7 = 35
\n\u2234 L.H.S = R.H.S<\/p>\n

ii) x3<\/sup> – y3<\/sup> = (x + y) (x2<\/sup> + xy + y2<\/sup>)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 x3<\/sup> – y3<\/sup> = (x – y) (x2<\/sup> + xy + y2<\/sup>)
\nL.H.S = x3<\/sup> – y3<\/sup>
\nR.H.S = (x – y) (x2<\/sup> + xy + y2<\/sup>)
\n= x (x2<\/sup> + xy + y2<\/sup>) -y (x2<\/sup> + xy + y2<\/sup>)
\n= x3<\/sup> + x2<\/sup>y + xy2<\/sup> – x2<\/sup>y – xy2<\/sup> – y3<\/sup>
\n= x3<\/sup> – y3<\/sup> = L.H.S
\nL.H.S = 33<\/sup> – 23<\/sup> = 27 – 8 = 19
\nR.H.S = (3 – 2) (32<\/sup> + 3 \u00d7 2 + 22<\/sup>)
\n= 1 \u00d7 (9 + 6 + 4) = 1 \u00d7 19 = 19
\n\u2234 L.H.S = R.H.S
\n\u0c2a\u0c48 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c4d\u0c2f\u0c35\u0c39\u0c30\u0c3f\u0c38\u0c4d\u0c24\u0c3e\u0c2e\u0c41.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2810.
\n9\u0c35 \u0c38\u0c2e\u0c38\u0c4d\u0c2f\u0c32\u0c4b \u0c2b\u0c32\u0c3f\u0c24\u0c3e\u0c32 \u0c06\u0c27\u0c3e\u0c30\u0c02\u0c17\u0c3e \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3e\u0c1f\u0c3f\u0c28\u0c3f \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c3f\u0c2d\u0c1c\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\ni) 27a3<\/sup> + 64b3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n27a3<\/sup> + 64b3<\/sup> = (3a)3<\/sup> + (4b)3<\/sup>
\n= (3a + 4b) {(3a)2<\/sup> – (3a) (4b) + (4b)2<\/sup>}
\n= (3a + 4b) (9a2<\/sup> – 12ab + 16b2<\/sup>)<\/p>\n

ii) 343y3<\/sup> – 1000
\n\u0c38\u0c3e\u0c27\u0c28.
\n343y3<\/sup> – 1000 = (7y)3<\/sup> – (10)3<\/sup>
\n= (7y – 10) [(7y)2<\/sup> + (7y) (10) + (10)2<\/sup>)
\n= (7y – 10) (49y2<\/sup> + 70y + 100)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2811.
\n\u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f 27×3 + y3 + z3 – 9xyz \u0c28\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c3e\u0c32\u0c41\u0c17\u0c3e \u0c35\u0c3f\u0c2d\u0c1c\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz
\n= (3x)3<\/sup> + (y)3<\/sup> + (z)3<\/sup> – 3 (3x) (y) (z)
\n= (3x + y + z) [(3x)2<\/sup> + y2<\/sup> + z2<\/sup> – (3x) (y) – (y) (z) – (z) (3x)
\n[\u2235 x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz \u2261 (x + y + z) – (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\n= (3x + y + 2) (9x2<\/sup> + y2<\/sup> + z2<\/sup> – 3xy – yz – zx)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2812.
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = \\(\\frac {1}{2}\\) (x + y + z) [(x + y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>] \u0c38\u0c30\u0c3f\u0c1a\u0c42\u0c21\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\n= \\(\\frac {1}{2}\\)(x + y + z) [(x – y)2<\/sup> + (y = z)2<\/sup> + (z = x)2<\/sup>]
\nR.H.S= \\(\\frac {1}{2}\\)(x + y + z) [(x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>]
\n= \\(\\frac {1}{2}\\)(x + y + z) [x2<\/sup> + y2<\/sup> – 2xy + y2<\/sup> + z2<\/sup> – 2yz + z2<\/sup> + x2<\/sup> – 2xz]
\n= \\(\\frac {1}{2}\\)(x + y + z) [2x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – 2xy – 2yz – 2zx]
\n= \\(\\frac {1}{2}\\)(x + y + z) (2) [x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx]
\n= (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\n= L.H.S<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2813.
\nx + y + z = 0 \u0c05\u0c2f\u0c3f\u0c24\u0c47 x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz \u0c05\u0c28\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 x + y + z = 0
\n\u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n(x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz -zx)
\n= x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz \u0c05\u0c28\u0c3f \u0c2e\u0c28\u0c15\u0c41 \u0c24\u0c46\u0c32\u0c41\u0c38\u0c41.
\nx + y + z = 0 \u0c28\u0c41 \u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c28\u0c02\u0c26\u0c41 \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c17\u0c3e,
\n0 \u00d7 (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx) = x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\n\u21d2 x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = 0
\n= x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2814.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c38\u0c3e\u0c32\u0c32\u0c4b \u0c18\u0c28\u0c3e\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c15\u0c41\u0c02\u0c21\u0c3e\u0c28\u0c47, \u0c2b\u0c32\u0c3f\u0c24\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\ni) (-10)3<\/sup> + 73<\/sup> + 33<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 (- 10)3<\/sup> + 73<\/sup> + 33<\/sup>
\n\u0c2d\u0c42\u0c2e\u0c41\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c02 = – 10 + 7 + 3 = 0
\n\u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n(x + y + z) (x2<\/sup> + y2 <\/sup>+ z2<\/sup> – xy – yz – zx)
\n= x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz \u0c28\u0c41\u0c02\u0c21\u0c3f
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz \u0c15\u0c3e\u0c35\u0c41\u0c28
\n\u2234 (-10)3<\/sup> + 73<\/sup> + 33<\/sup> = 3 (-10) \u00d7 (7) \u00d7 3
\n= -630<\/p>\n

ii) (28)3<\/sup> + (- 15)3<\/sup> + (-13)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 (28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\n\u0c2d\u0c42\u0c2e\u0c41\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c02 = 28 + (-15) + (-13) = 0
\n\u0c38\u0c30\u0c4d\u0c35\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx) \u0c28\u0c41\u0c02\u0c21\u0c3f x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz \u0c15\u0c3e\u0c35\u0c41\u0c28
\n\u2234 (28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\n= 3 \u00d7 28 \u00d7 (-15) \u00d7 (13)
\n= 16380<\/p>\n

iii) \\(\\left(\\frac{1}{2}\\right)^{3}+\\left(\\frac{1}{3}\\right)^{3}-\\left(\\frac{5}{6}\\right)^{3}\\)
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 \\(\\left(\\frac{1}{2}\\right)^{3}+\\left(\\frac{1}{3}\\right)^{3}-\\left(\\frac{5}{6}\\right)^{3}\\)\u0c28\u0c41 \\(\\left(\\frac{1}{2}\\right)^{3}+\\left(\\frac{1}{3}\\right)^{3}+\\left(\\frac{-5}{6}\\right)^{3}\\) \u0c17\u0c3e \u0c35\u0c4d\u0c30\u0c3e\u0c2f\u0c35\u0c1a\u0c4d\u0c1a\u0c41\u0c28\u0c41.
\n\"AP<\/p>\n

iv) (0.2)3<\/sup> – (0.3)3<\/sup> + (0.1)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c2e\u0c41 (0.2)3<\/sup> – (0.3)3<\/sup> + (0.1)3<\/sup> \u0c28\u0c41
\n= (0.2)3<\/sup> + (-0.3)3<\/sup> + (0.1)3<\/sup> \u0c17\u0c3e \u0c35\u0c4d\u0c30\u0c3e\u0c2f\u0c35\u0c1a\u0c4d\u0c1a\u0c41\u0c28\u0c41.
\n\u0c2d\u0c42\u0c2e\u0c41\u0c32 \u0c2e\u0c4a\u0c24\u0c4d\u0c24\u0c02 = 0.2 – 0.3 + 0.1 = 0
\n\u2234 (0.2)3<\/sup> + (-0.3)3<\/sup> + (0.1)3<\/sup>
\n= 3 \u00d7 (0.2) (-0.3) (0.1) = – 0.018<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2815.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c26\u0c40\u0c30\u0c4d\u0c18\u0c1a\u0c24\u0c41\u0c30\u0c38\u0c4d\u0c30\u0c3e\u0c32 \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c3e\u0c32\u0c15\u0c41 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c38\u0c2e\u0c3e\u0c38\u0c3e\u0c32\u0c28\u0c41 \u0c2c\u0c1f\u0c4d\u0c1f\u0c3f \u0c2a\u0c4a\u0c21\u0c35\u0c41, \u0c35\u0c46\u0c21\u0c32\u0c4d\u0c2a\u0c41\u0c32\u0c15\u0c41 \u0c24\u0c17\u0c3f\u0c28 \u0c05\u0c28\u0c41\u0c15\u0c42\u0c32 \u0c15\u0c4a\u0c32\u0c24\u0c32 \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c24\u0c46\u0c32\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c2e\u0c41 = 4a2<\/sup> + 4a – 3
\n= 4a2<\/sup> + 6a – 2a – 3
\n= 2a (2a + 3) -1 (2a + 3)
\n= (2a + 3) (2a – 1)
\n\u2234 \u0c2a\u0c4a\u0c21\u0c35\u0c41 = (2a + 3); \u0c35\u0c46\u0c21\u0c32\u0c4d\u0c2a\u0c41 = (2a – 1)<\/p>\n

ii) 25a2<\/sup> – 35a + 12
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c35\u0c48\u0c36\u0c3e\u0c32\u0c4d\u0c2f\u0c2e\u0c41 = 25a2<\/sup> – 35a + 12
\n= 25a2<\/sup> – 20a – 15a + 12
\n= 5a (5a – 4) – 3 (5a – 4)
\n= (5a – 4) (5a – 3)
\n\u2234 \u0c2a\u0c4a\u0c21\u0c35\u0c41 = (5a – 3); \u0c35\u0c46\u0c21\u0c32\u0c4d\u0c2a\u0c41 = (5a – 4)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2816.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c26\u0c40\u0c30\u0c4d\u0c18\u0c18\u0c28 \u0c18\u0c28\u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c3e\u0c32\u0c15\u0c41 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c38\u0c2e\u0c3e\u0c38\u0c3e\u0c32\u0c28\u0c41 \u0c2c\u0c1f\u0c4d\u0c1f\u0c3f \u0c26\u0c40\u0c30\u0c4d\u0c18\u0c18\u0c28\u0c02 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c05\u0c28\u0c41\u0c15\u0c42\u0c32 \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c41 \u0c24\u0c46\u0c32\u0c2a\u0c02\u0c21\u0c3f.
\ni) 3x3<\/sup> – 12x
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c38\u0c2e\u0c18\u0c28\u0c2a\u0c41 \u0c18\u0c28\u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41 = 3x3<\/sup> – 12x = 3x (x2<\/sup> – 4)
\n= 3x (x + 2) (x – 2)
\n{\u2235 a2<\/sup> – b2<\/sup> = (a + b) (a – b)}
\n\u2234 \u0c38\u0c2e\u0c18\u0c28\u0c2a\u0c41 \u0c05\u0c28\u0c41\u0c15\u0c42\u0c32 \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e
\n3x, (x + 2) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (x – 2)<\/p>\n

ii) 12y2<\/sup> + 8y – 20
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28 \u0c18\u0c28\u0c2a\u0c41 \u0c18\u0c28\u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02
\n= 12y2<\/sup> + 8y – 20
\n= 4 (3y2<\/sup> + 2y – 5)
\n= 4 [3y2<\/sup> + 5y – 3y – 5]
\n= 4 [y (3y + 5) – 1 (3y + 5)]
\n= 4 (3y + 5) (y – 1)
\n\u2234 \u0c38\u0c2e\u0c18\u0c28\u0c2a\u0c41 \u0c05\u0c28\u0c41\u0c15\u0c42\u0c32 \u0c15\u0c4a\u0c32\u0c24\u0c32\u0c41 \u0c35\u0c30\u0c41\u0c38\u0c17\u0c3e
\n4, (3y + 5) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 (y – 1)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c2817.
\n2 (a2<\/sup> + b2<\/sup>) = (a + b)2<\/sup> \u0c05\u0c2f\u0c3f\u0c28 a = b \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\n\u0c26\u0c24\u0c4d\u0c24\u0c3e\u0c02\u0c36\u0c02 2 (a2<\/sup> + b2<\/sup>) = (a + b)2<\/sup>
\n2a2<\/sup> + 2b2<\/sup>= a2<\/sup> + 2ab + b2<\/sup>
\n2a2<\/sup> – a2<\/sup> + 2b2<\/sup> – b2<\/sup> = 2ab
\na2<\/sup> + b2<\/sup> = 2ab
\na = b \u0c05\u0c2f\u0c3f\u0c28\u0c2a\u0c41\u0c21\u0c41 \u0c2e\u0c3e\u0c24\u0c4d\u0c30\u0c2e\u0c47 \u0c07\u0c26\u0c3f \u0c38\u0c3e\u0c27\u0c4d\u0c2f\u0c2e\u0c41.<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 9th Class Maths Solutions Chapter 2 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15 \u0c35\u0c3f\u0c2d\u0c1c\u0c28 Ex 2.5 Textbook Exercise Questions and Answers. AP State Syllabus 9th Class Maths Solutions 2nd Lesson \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15 \u0c35\u0c3f\u0c2d\u0c1c\u0c28 Exercise 2.5 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c281. \u0c24\u0c17\u0c3f\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c09\u0c2a\u0c2f\u0c4b\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c32\u0c2c\u0c4d\u0c26\u0c3e\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c41\u0c2e\u0c41. i) (x + 5) (x + 2) \u0c38\u0c3e\u0c27\u0c28. (x + 5) (x + 2) …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29411"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=29411"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29411\/revisions"}],"predecessor-version":[{"id":29416,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/29411\/revisions\/29416"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=29411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=29411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=29411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}