`a)` Với `x \ne +-1;x \ne -1/2` có:
`A=(1/[x-1]+x/[x^2-1]).[x^2-x]/[2x+1]`
`A=[x+1+x]/[(x-1)(x+1)].[x(x-1)]/[2x+1]`
`A=[2x+1]/[(x-1)(x+1)].[x(x-1)]/[2x+1]=x/[x+1]`
`b)` Thay `x=9` (t/m đk) vào `A` thu gọn có:
`A=9/[9+1]9/10`
`c)` Với `x \ne +-1;x \ne -1/2` có:
`A=x/[x+1]=[x+1-1]/[x+1]=1-1/[x+1]`
Để `A in ZZ` thì `1-1/[x+1] in ZZ`
`=>1/[x+1] in ZZ`
`=>x+1 in Ư_1`
Mà `Ư_1={+-1}`
`=>{(x+1=1<=>x=0),(x+1=-1<=>x=-2):}` `(t//m x \ne +-1;x \ne -1/2)`
Vậy `x in {0;-2}` thì `A in ZZ`
a. ĐKXĐ : \(x\ne\left\{\pm1;-\dfrac{1}{2}\right\}\)
\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{x^2-1}\right).\dfrac{x^2-x}{2x+1}\)
\(=\left[\dfrac{x+1+x}{\left(x+1\right)\left(x-1\right)}\right].\dfrac{x\left(x-1\right)}{2x+1}\)
\(=\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\)
\(=\dfrac{x\left(2x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(2x+1\right)}=\dfrac{x}{x+1}\)
Vậy : \(A=\dfrac{x}{x+1}\)
b. Do x = 9 tmđk nên thay x = 9 vào biểu thức A :
\(A=\dfrac{9}{9+1}=\dfrac{9}{10}\)
Vậy : Khi x = 9 thì \(A=\dfrac{9}{10}\)
c. Ta có : \(A=\dfrac{x}{x+1}=\dfrac{x+1-1}{x+1}=1-\dfrac{1}{x+1}\)
A nguyên khi \(1⋮\left(x+1\right)\Leftrightarrow\left(x+1\right)\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy : A nguyên khi \(x=\left\{-2;0\right\}\).