[" 37."((2x+1)!)/((x+2)!)times((x-1)!)/((2x-1)!)=(3)/(5);(x in1N)" र्ल,"x],[" Эर मान २(ठ ? "]

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

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

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

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

(1-2x+x^(2))/(1-x^(3))times(1+x+x^(2))/(1+x)

y=(e^(xx)x sin xx x(x+1)^(2))/((2x+1)^(3)x^(5)),( find )(dy)/(dx)

solve for x,4^(x-1)xx(0.5)^(3-2x)=((1)/(8))^(-x)

Solve for x : 4^(x-1)xx(0.5)^(3-2x)=((1)/(8))^(-x)

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 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)