Question

Fie expresia: $\bold{E(x)=(\frac{4}{x-1}+ \frac{13-5x}{1-x^2}- \frac{2x+6}{x^2+4x+3}): \frac{1}{x+1}}$
a)Determinati valorile lui x pentru care E(x) are sens.
b)Aduceti expresia la forma cea mai simpla.

Answer 1

$a) x-1=0=>x=1 \\ 1- x^{2} =0 => x^{2} =1 =>x=1~sau~x=-1 \\ x^{2} +4x+3=0 <=> x^{2} +x+3x+3=0<=> \\ <=>x(x+1)+3(x+1)=0 <=>(x+1)(x+3)=0=> \\ => x=-1~sau~x=-3.$
Fractia nu este definita pentru x∈{-3;-1;1}.
D=R-{-3;-1;1}.

$b)E(x)=[ \frac{4}{x-1}+ \frac{13-5x}{(1-x)(1+x)}- \frac{2(x+3)}{(x+1)(x+3)}]*(x+1) = \\ =[ \frac{4}{x-1}- \frac{13-5x}{(x-1)(x+1)}- \frac{2}{x+1} ]*(x+1)= \\ =[ \frac{4(x+1)}{(x-1)(x+1)} }- \frac{13-5x}{(x-1)(x+1)}- \frac{2(x-1)}{(x-1)(x+1)}]*(x+1)= \\ = \frac{4(x+1)-13+5x-2(x-1)}{(x-1)(x+1)}*(x+1) = \\ = \frac{4x+4-13+5x-2x+2}{x-1}= \\ = \frac{7x-7}{(x-1)}= \\ = \frac{7(x-1)}{x-1}= \\ =7.$