\(A=[\dfrac{2}{\left(x+1\right)^3}.\dfrac{1+x}{x}+\left(\dfrac{1}{\left(x+1\right)^2}.\dfrac{1+x^2}{x^2}\right)].\dfrac{x^3}{x-1}=\left(\dfrac{2+2x}{x\left(x+1\right)^3}+\dfrac{1+x^2}{x^2}\right).\dfrac{x^3}{x-1}=\dfrac{2x+2x^2+\left(1+x^2\right)\left(x+1\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{2x\left(1+x\right)+\left(1+x^2\right)\left(x+1\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{\left(x+1\right)\left(2x+1+x^2\right)}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{\left(x+1\right)^3}{x^2\left(x+1\right)^3}.\dfrac{x^3}{x-1}=\dfrac{x\left(x+1\right)}{x-1}=\dfrac{x^2+x}{x-1}\)
ý a có bn lm rồi, mk lm ý b,c thôi nhé
b/ A < 1 \(\Leftrightarrow\dfrac{x^2+x}{x-1}< 1\)
\(\Leftrightarrow x^2+x< x-1\)
\(\Leftrightarrow x^2+x-x+1< 0\)
\(\Leftrightarrow x^2+1< 0\)
\(\Leftrightarrow x^2< -1\) (vô lí)
Vậy k có gt nào của x t/m
c/ \(\dfrac{x^2+x}{x-1}=\dfrac{x^2+x-2+2}{x-1}=\dfrac{\left(x+2\right)\left(x-1\right)+2}{x-1}\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)}{x-1}+\dfrac{2}{x-1}=x+2+\dfrac{2}{x-1}\)
Để A \(\in\) Z <=> \(\dfrac{2}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(2\right)\)
\(\Leftrightarrow x-1=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x=\left\{-1;0;2;3\right\}\)
Vậy....
