Bài 1:
a) ĐKXĐ: \(x\notin\left\{1;-1;-\dfrac{1}{2}\right\}\)
Ta có: \(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}-\dfrac{5}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(=\dfrac{2\left(x-1\right)-\left(x+1\right)+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(=\dfrac{2x-2-x-1+5}{2x+1}\)
\(=\dfrac{x+2}{2x+1}\)
b) Để A=3 thì \(x+2=3\left(2x+1\right)\)
\(\Leftrightarrow x+2=6x+3\)
\(\Leftrightarrow-5x=1\)
hay \(x=-\dfrac{1}{5}\)(thỏa ĐK)
c) Ta có: \(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào A, ta được:
\(A=\dfrac{0+2}{2\cdot0+1}=\dfrac{2}{1}=2\)
d) Để A nguyên thì \(x+2⋮2x+1\)
\(\Leftrightarrow2x+1+3⋮2x+1\)
\(\Leftrightarrow3⋮2x+1\)
\(\Leftrightarrow2x+1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow2x\in\left\{0;-2;2;-4\right\}\)
\(\Leftrightarrow x\in\left\{0;-1;1;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2\right\}\)
