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

(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=

(x+1)/(2)+(x-1)/(3)=(5)/(12)(x-2)

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

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)

Let f(x)=|[1,1,1] , [3-x,5-3x^2,3x^3-1] , [2x^2-1,3x^5-1,7x^8-1]| then the equation of f(x)=0 has

Let f(x)=|[1,1,1] , [3-x,5-3x^2,3x^3-1] , [2x^2-1,3x^5-1,7x^8-1]| then the equation of f(x)=0 has

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

The equation |{:((1+x)^(2),(1-x)^(2),-(2+x^(2))),(2x+1,3x,1-5x),(x+1,2x,2-3x):}|+|{:((1+x)^(2),2x+1,x+1),((1-x)^(2),3x,2x),(1-2x,3x-2,2x-3):}|=0