`7(x-2y)^2-25(x-2y)+12`

Factorise : (x-2y)^(2) - 12(x-2y) + 32

Factorise: 25(x+y)^(2)- 36 (x-2y)^(2).

The circles x^2+y^2-12x-12y=0 and x^2+y^2+6x+6y=0 :

Find the product. (3x-5y-4)(9x^2+25y^2+15xy-20y+12x+16)

The figure formed by the pairs of lines 12x^(2)+7xy-12y^(2)=0 and 12x^(2)+7xy-12y^(2)+10x+5y+2=0 is a

The four lines given by the equations 12 x^(2)+7 x y-12 y^(2)=0 and 12 x^(2)+7 x y-12 y^(2)-x+7 y-1=0 lie along the sides of a

If x^(2)+y^(2)=25, xy=12 then x=

If x=5 and y=-2 , then find the value of (i) (8x-2y+5)/(2x+4ty+12) " "(ii) (5x-6y-7)/(3x+6y+12)

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0