(v)(2x-1)(x-3)=(x+5)(x-1)

(2x-1)/((x-1)(2x+3))=1/(5(x-1))-k/(5(2x+3)) , then k =

((2x-1)/((x-1)(2x+3))=(1)/(5(x-1))+(k)/(5(2x+3))rArr k=

Check whether the following are quadratic equations : (1) (x-1)^(2)=2(x-3) (2) x^(2)-2x=(-2)(3-x) (3) (x-2)(x+1)=(x-1)(x+3) (4) (x-3)(2x+1)=x(x+5) (5) (2x-1)(x-3)=(x+5)(x-1) (6) x^(2)+3x+1=(x-2)^(2) (7) (x+2)^(3)=2x(x^(2)-1) (8) x^(3)-4x^(2)-x+1=(x-2)^(3)

(3)/(x+1)-(2)/(x-1)=(5)/(x^(2)-1)

(v) (2x-5)(3x+7) - (3x-2)(x+1)

Expand using binomial theorem: (i) (1-2x)^(4) " " (ii) (x+2y)^(5) (iii) (x-(1)/(x))^(6) " "(iv) ((2x)/(3)=(3)/(2x))^(5) (v) (x^(2) +(2)/(x))^(6)" "(vi) (1+(1)/(x^(2)))^(4)

Solve ((2x+1)!)/((x+2)!)xx((x-1)!)/((2x-1)!)=(3)/(5)(x in N)

The equation is true for (12x+1)/(4)=(15x-1)/(5)+(2x-5)/(3x-1)

(i) (2x+3) (3x-5) (ii) x(1+x)^(3) (iii) (sqrtx + 1/x) (x -1/sqrtx) (iv) (x-1/x)^(2) (v) (x^(2) + 1/x^(2))^(3) (vi) (2x^(2) +5x-1) (x-3)