Giải thích các bước giải:
$P=(\dfrac{x+2}{x^2+2x+1}-\dfrac{x-2}{x^2-1}):\dfrac{x}{x+1}$
$\to P=(\dfrac{x+2}{(x+1)^2}-\dfrac{x-2}{(x-1)(x+1)}).\dfrac{x+1}{x}$
$\to P=(\dfrac{x+2}{x+1}-\dfrac{x-2}{x-1}).\dfrac{1}{x}$
$\to P=\dfrac{(x+2)(x-1)-(x+1)(x-2)}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2x}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2}{(x+1)(x-1)}$
b.Để $P=4\to \dfrac{2}{(x+1)(x-1)}=4\to \dfrac{2}{x^2-1}=4\to x^2-1=\dfrac{1}{2}\to x^2=\dfrac{3}{2}\to x=\pm\sqrt{\dfrac{3}2}$
c.$x=5\to P=\dfrac{2}{(5+1)(5-1)}=\dfrac{1}{12}$
d.Để $P > 1\to \dfrac{2}{(x+1)(x-1)}>1\to \dfrac{2}{x^2-1}>1\to 0< x^2-1<2$
$\to -\sqrt{3}<x<-1 $ hoặc $1<x<\sqrt{3}$
