" जाँच कीजिए कि "(2x-1)(x-3)=(x+5)(x-1)" द्विधात समीकरण है। "

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

2x-1/3=1/5-x

solve for x (2x-5)/(3x-1)=(2x-1)/(3x+2)

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)

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

निम्न समीकरण को हल कीजिए - tan^(-1)(x-1)+tan^(-1)(x)+tan^(-1)(x+1)=tan^(-1) 3 x

If (3x+1)^3+(x-3)^3+(4-2x)^3+ 6 (3x+1) (x -3) (x -2)=0 , then x is equal to: यदि (3x+1)^3+(x-3)^3+(4-2x)^3+ 6 (3x+1) (x -3) (x -2)=0 है तो x बराबर है:

If x≠ − 1 , 2 and 5, then the simplified value of {(2(x^3 -8))/(x^2-x -2) xx (x^2+ 2x +1)/(x^2 -4x-5) ÷ (x^2 +2x+4)/(3x-15)} is equal to: यदि x≠ − 1 , 2 और 5 है तो {(2(x^3 -8))/(x^2-x -2) xx (x^2+ 2x +1)/(x^2 -4x-5) ÷ (x^2 +2x+4)/(3x-15)} का सरलीकृत मान बराबर हैः

Check whether the following are quadratic equation (i) (x+1)^2=2(x-3) (ii) x^2-2x=(-2)(3-x) (iii) (x-2)(x+1)=(x-1)(x+3) (iv) (x-3)(2x+1)=x(x+5) (v) (2x-1)(x-3)=(x+5)(x-1) (vi) x^2+3x+1=(x-2)^2 (vii) (x+2)^3=2x(x^2-1) (viii) x^3-4x^2 – x + 1 = (x – 2)^3

Check whether the following are quadratic equation (i) (x+1)^2=2(x-3) (ii) x^2-2x=(-2)(3-x) (iii) (x-2)(x+1)=(x-1)(x+3) (iv) (x-3)(2x+1)=x(x+5) (v) (2x-1)(x-3)=(x+5)(x-1) (vi) x^2+3x+1=(x-2)^2 (vii) (x+2)^3=2x(x^2-1) (viii) x^3-– 4x^2 – x + 1 = (x – 2)^3