e) Để A nguyên thì \(-3⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;2;-4\right\}\)
a) Ta có: \(A=\left(\dfrac{x^2-2x+1}{x^2-1}-\dfrac{2x^2-5x}{x^2+x}\right)\cdot\dfrac{3}{x-4}\)
\(=\left(\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(2x-5\right)}{x\left(x+1\right)}\right)\cdot\dfrac{3}{x-4}\)
\(=\dfrac{x-1-2x+5}{x+1}\cdot\dfrac{3}{x-4}\)
\(=\dfrac{-x+4}{x+1}\cdot\dfrac{3}{x-4}\)
\(=\dfrac{-3\left(x-4\right)}{\left(x-4\right)\left(x+1\right)}=\dfrac{-3}{x+1}\)
b) Thay x=-4 vào A, ta được:
\(A=\dfrac{-3}{-4+1}=\dfrac{-3}{-3}=1\)
c) Để A=-3 thì x+1=1
hay x=0(loại)
d) Để A>0 thì x+1<0
hay x<-1
