{(y=(1)/(2)x-2), (y=-x^(2)+1):} যদি (a, b) উপরের সমীকরণের সিস্টেমের একটি সমাধান হয়, যা...

If y=(x^(2)+x+1)/(x^(2)-x+1)," then "(x^(2)-x+1)^(2)y_(1)=

{:("Column" A ,, "Column" B), (225x^(2) - 625 y^(2) = ,, (a) 25(x-2) (x-2)), (x^(2) - x - y - y^(2) = ,, (b) 25(3x- 5y) (3x + 5y)), (x^(2) - x - y^(2) + y = ,, (x + y) (x - y- 1)), (25x^(2) - 100 x + 100 = ,, (d) (x - y) (x + y -1)), (,,(e) (x + y) (x + y - 1)):}

(x + 2) (x + 1) = (x -2) (x -3) (y+3)(y+2)=(y-1)(y-2)

If y=tan^(-1)x, then (1+x^(2))y_(2)+2xy_(1)= (a) y(b)0(c)1(d)2

|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|=|(a_1,b_1,1),(a_2,b_2,1),(a_3,b_3,1)| then the two triangles with vertices (x_(1), y_(1)), (x_(2), y_(2)), (x_(3), y_(3)) and (a_(1), b_(1)), (a_(2), b_(2)), (a_(3), b_(3)) are

2x + 3y = 7 , (a+b + 1 )x + (a + 2b + 2) y = 4 (a + b ) + 1