Question

Answer 1

a.

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

b.

\(x=3\Rightarrow A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)

\(x=-\dfrac{3}{2}\Rightarrow A=\dfrac{-\dfrac{3}{2}-1}{-\dfrac{3}{2}+1}=5\)

c.

\(A=\dfrac{x-1}{x+1}=\dfrac{x+1-2}{x+1}=1-\dfrac{2}{x+1}\)

A nguyên khi \(\dfrac{2}{x+1}\) nguyên \(\Rightarrow x+1=Ư\left(2\right)\)

\(\Rightarrow x+1=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow x=\left\{-3;-2;0;1\right\}\)

Kết hợp ĐKXĐ \(\Rightarrow x=\left\{-3;-2;0\right\}\)

Answer 2

a) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

\(=\dfrac{x-1}{x+1}\)

b) Thay `x=3` vào A ta có: \(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

Thay `x=-3/2` vào A ta có: \(A=\left(-\dfrac{3}{2}-1\right):\left(-\dfrac{3}{2}+1\right)=-\dfrac{5}{2}:\dfrac{-1}{2}=5\)

c) Ta có: \(A=\dfrac{x-1}{x+1}=\dfrac{x+1-2}{x+1}=1-\dfrac{2}{x+1}\)

Để A nguyên thì x + 1∈ Ư(2) = {1; -1; 2; -2}

⇒ x ∈ {0; -2; 1; -3}