y=(x^(2)-2x+5)/(x^(2)+2x+5)

If (3x-5y)/(3x+5y) = 1/2 . Then find the value of (3x^(2)-5y^(2))/(3x^(2) + 5y^(2)) .

If x ^(2) - 2x ^(2) y ^(2) + 5x + y-5 =0 and y (1) then y '' (1)=

y= (2x+5)/(3x-2)

If x:y::5:2, then (x ^(2) +y)/(x + y ^(2)) =?

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))

(x^(2)+4x)^(2)-5(x^(2)+4x)-y(x^(2)+4x)+5y

Factorisation by grouping the terms: (x^(2)+3x)^(2)-5(x^(2)+3x)-y(x^(2)+3x)+5y

3 - [x - {2y - (5x + y - 3) + 2x^(2)} - (x^(2) - 3y)]

x"@"y=(x+y)^(2) x#y=(x-y)^(2) (2" @ "2)+(3#3)-(4"@"4)+(5#5) is equal to :