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Aratati ca ([tex] \frac{2}{x+1} [/tex] + [tex] \frac{x}{1- x^{2} } [/tex] - [tex] \frac{3x+6}{ x^{2}+x-2} [/tex]) : [tex] \frac{1}{1-x} [/tex]= [tex] \frac{2x+5}{x+1} [/tex]

Răspuns :

[tex]E= (\frac{2}{x+1}+ \frac{x}{(1-x)(1+x)}- \frac{3(x+2)}{(x+2)(x-1)}): \frac{1}{1-x} \\ E= (\frac{2}{x+1}+ \frac{x}{(1-x)(1+x)}- \frac{3}{(x-1)}): \frac{1}{1-x} \\ E= (\frac{2}{x+1}+ \frac{x}{(1-x)(1+x)}+ \frac{3}{(1-x)}): \frac{1}{1-x} \\ E= \frac{2-2x+x+3x+3}{(1-x)(1+x)}\cdot (1-x) \\ E= \frac{2x+5}{1+x}=\frac{2x+5}{x+1} [/tex]