Question

[hà nguyễn]

làm phần a bài 8 hộ mình với ạ 

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 8:

a: Ta có: \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\cdot\dfrac{x^4-2x^2+1}{2}\)

\(=\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{2}\)

\(=\dfrac{x^2-x-2-x^2-x-2}{1}\cdot\dfrac{x-1}{2}\)

\(=\dfrac{-2x\cdot\left(x-1\right)}{2}=-x\left(x-1\right)\)

Answer 2

[Lấp La Lấp Lánh]

Bài 8:

a) \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\dfrac{x^4-2x^2+1}{2}\left(đk:x\ne1,x\ne-1\right)\) 

\(=\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{\left(x^2-1\right)^2}{2}=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}.\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{2}=\dfrac{-2x\left(x-1\right)}{2}=-x^2+x\)

b) \(x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)\(\Leftrightarrow x=2\)(do đkxđ của A là \(x\ne1\))

\(A=-x^2+x=-2^2+2=-2\)

c) Do \(A=-x^2+x\in Z\forall x\in Z\)

\(\Rightarrow A\in Z\Leftrightarrow x\in Z\)