[" 15."sqrt(2)(4x^(2)-3y^(2)):(2x^(2)+5y^(2))=12:19" हो,तो "x:y=?],[[" (a) "2:3," (b) "1:2," (c) "3:2],[" (t) "1,2-4x," Ho - so "x*y=?]]

If (4x^2-3y^2):(2x^2+5y^2)=12:19 then x:y=

3x^(2) + 8x +4=0, 4y^(2)-19y + 12=0

If (3x-y):x(4x+y)=5:12 then find (x^(2)+y^(2)):(x+y^(2))

Solve for (x - 1)^(2) and (y + 3)^(2) , 2x^(2) - 5y^(2) - x - 27y - 26 = 3(x + y + 5) and 4x^(2) - 3y^(2) - 2xy + 2x - 32y - 16 = (x - y + 4)^(2) .

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

Multiply: 9x^(2)+25y^(2)+15xy+12x-20y+16 by 3x-5y+4

The radical centre of the circles x^(2)+y^(2)=9 , x^(2)+y^(2)-2x-2y-5=0 , x^(2)+y^(2)+4x+6y-19=0 is A) (0,0) (B) (1,1) (C) (2,2) (D) (3,3)

Given x, y in R , x^(2) + y^(2) gt 0 . Then the range of (x^(2) + y^(2))/(x^(2) + xy + 4y^(2)) is (a) ((10 - 4 sqrt(5))/(3),(10 + 4 sqrt(5))/(3)) (b) ((10 - 4 sqrt(5))/(15),(10 + 4 sqrt(5))/(15)) (c) ((5- 4 sqrt(5))/(15),(5 + 4 sqrt(5))/(15)) (d) ((20- 4 sqrt(5))/(15),(20 + 4 sqrt(5))/(15))