Question

Ma poate ajuta cineva va rog frumos ?

Answer 1

Răspuns:

Explicație pas cu pas:

a)[2x/(1+x)+2/(x-1)+4x/(x²-1)]·[2x/(x+1)+2/(x-1)-4x/(x²-1)]=

[2x(x-1)+2(1+x)+4x]/(x+1)(x-1)·[2x(x-1)+2(x+1)-4x]/(x+1)(x-1)=

(2x²-2x+2+2x+4x)/(x+1)(x-1)·(2x²-2x+2x+2-4x)//(x+1)(x-1)=

(2x²+4x+2)/(x+1)(x-1)·(2x²-4x+2)/(x+1)(x-1)=

2(x²+2x+1)/(x+1)(x-1)·2(x²-2x+1)/(x+1)(x-1)=2(x+1)²/(x+1)(x-1)·2(x-1)²/(x+1)(x-1)=

2(x+1)/(x-1)·2(x-1)/(x+1)=4

b) [x/(x-1)+2/(x-2)-2/(x²-3x+2)]:(x²+3x+2)/(x²+2x-3)=

[x(x-2)+2(x-1)-2]/(x-1)(x-2)]:[(x+1)²+x+1]/(x²-1+2x-2)=

[(x²-2x+2x-2-2)/(x-1)(x-2)]:[(x+1)(x+1+1)/[(x-1)(x+1)+2(x-1)]=

[(x²-4)/(x-1)(x-2)]:[(x+1)(x+2)/(x-1)(x+1+2)]=

[(x-2)(x+2)/(x-1)(x-2)]:[(x+1)(x+2)/(x-1)(x+3)]=[(x+2)/(x-1)]·[(x-1)(x+3)/(x+1)(x+2)]=

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

c) [(x+1)/(2x+3)-(x-2)/(2x-3)+(x-4)/(4x²-9)]·(2x²-x-3)/(x²-1)=

{[(x+1)(2x-3)-(x-2)(2x+3)+x-4]/(2x+3)(2x-3)}·[(2x²-2-x-1]/(x+1)(x-1)=

[(2x²-3x+2x-3-2x²-3x+4x+6+x-4)/(2x+3)(2x-3)]·[2(x²-1)-(x+1)]/(x+1)(x-1)=

](x-1)/(2x+3)(2x-3)]·[2(x+1)(x-1)-(x+1)]/(x+1)(x-1)=

[(x-1)/(2x+3)(2x-3)]·(x+1)(2x-2-1)/(x+1)(x-1)=

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

d) [4-8x/(x+2)]:[x/(x-2)+4/(x+2)-(2x+4)/(x²-4)]=

[4(x+2)-8x]/(x+2):[(x(x+2)+4(x-2)-2x-4]/(x+2)(x-2)=

(4x+8-8x)/(x+2):[(x²+2x+4x-8-2x-4)/(x+2)(x-2)=

(-4x+8)/(x+2):[(x²+4x-12)/(x+2)(x-2)=-4(x-2)/(x+2):(x²-4+4x-8)/(x+2)(x-2)=

-4(x-2)/(x+2):[(x+2)(x-2)+4(x-2)]/(x+2)(x-2)=

-4(x-2)/(x+2):(x-2)(x+2+4)/(x+2)(x-2)=[-4(x-2)/(x+2)]·(x+2)(x-2)/(x-2)(x+6)=

-4(x-2)(x+2)(x-2)/(x+2)(x-2)(x+6)=-4(x-2)/(x+6)