गुणनखंड: (2x +5y)^2-1 | कक्षा 8 | फ़ैक्टरीकरण | गणित | संशय

Factorise: (2x +5y)^2-1

8x^2y^2-x^5

2x + 5y = 1 , 2x + 3y = 3 .

2x - 5y + 4= 0 , 2x + y - 8 = 0

{:(2x - 5y + 4 = 0),(2x + y - 8 = 0):}

2x+3y=8 4x-5y=-6

The equation of the circle having the intercept on the line y+2x=0 by the circle x^2 + y^2 + 4x + 6y = 0 as a diameter is : (A) 5x^2 + 5y^2 - 8x + 16y =0 (B) 5x^2 + 5y^2 + 8x - 16y =0 (C) 5x^2 + 5y^2 - 8x - 16y =0 (D) 5x^2 + 5y^2 + 8x+- 16y =0

4x+2y = 8.5, 2x + 4y = 9.5

Find each of the following products: (i) (x + 3) (x - 3) (ii) (2x + 5)(2x - 5) (ii) (8 + x)(8 - x) (iv) (7x + 11y) (7x - 11y) (v) (5x^(2) + (3)/(4) y^(2)) (5x^(2) - (3)/(4) y^(2)) (vi) ((4x)/(5) - (5y)/(3)) ((4x)/(5) + (5y)/(3)) (vii) (x + (1)/(x)) (x - (1)/(x)) (viii) ((1)/(x) + (1)/(y)) ((1)/(x) - (1)/(y)) (ix) (2a + (3)/(b)) (2a - (3)/(b))

Add: 4x^(2) - 7xy + 4y^(2) - 3, 5 + 6y^(2) - 8xy + x^(2) and 6 - 2xy + 2x^(2) - 5y^(2)