Question

E(x)= (1+$\frac{2-x}{x+1}$ ) : $\frac{x-1}{(2x+1) ^{2}- (x+2) ^{2} }$, unde x esre nr. real , x≠1 si x≠-1.Aratati ca E(x)=9 ,pentru oricare x nr. real, x≠1 si x≠-1
Explicati va rog si modul de calcul!

Answer 1

=[1+(2-x)/(x+1)]:[(x-1)/(4x²+2x+2x+1-x²-2x-2x-4)=
=[(x+1)+(2-x)]/(x+1):[(x-1)/(3x²-3)]=
=[3/(x+1)]:[(x-1)/3(x-1)(x+1)] simplificam in paranteza mare cu x-1=
=3/(x+1):1/3(x+1)=
=3/(x+1)·3(x+1)/1=simplificam cu x+1 si cu 3=
=3·3=9

Answer 2

E(x)=(1+$\frac{2-x}{x+1}$):$\frac{x-1}{ (2x+1)^{2} - (x+2)^{2} }$=

=$\frac{x+1+2-x}{x+1}$ * $\frac{ (2x+1)^{2} - (x+2)^{2}{x-1} }$=

=$\frac{3}{x+1}$ * $\frac{4 x^{2} +4x+1-( x^{2} +4x+4)}{x-1}$=

=$\frac{3}{x+1}$ * $\frac{3 x^{2} -3}{x-1}$=

=$\frac{3}{x+1}$ * $\frac{3( x^{2} -1)}{x-1}$=

=$\frac{3}{x+1}$ * $\frac{3(x -1)(x+1)}{x-1}$=

=9