a) \(ĐKXĐ:x\ne\pm1\)

\(A=\frac{x^3-2x^2+x}{x^2-1}\)

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

\(\Leftrightarrow A=\frac{x^2-x}{x+1}\)

b) Để A có giá trị nguyên

\(\Leftrightarrow\frac{x^2-x}{x+1}\inℤ\)

\(\Leftrightarrow x^2-x⋮x+1\)

\(\Leftrightarrow x^2-x-2+2⋮x+1\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)+2⋮x+1\)

\(\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Ta sẽ loại các giá trị ktm

\(\Leftrightarrow x\in\left\{-2;0;-3\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;0;-3\right\}\)

a: ĐKXĐ: x<>1; x<>-1

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

c: Để A nguyên thì x+1-2 chia hết cho x+1

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

=>\(x\in\left\{0;-2;-3\right\}\)

a) Biểu thức A xác định khi \(\hept{\begin{cases}x+1\ne0\\x^2-1\ne0\end{cases}\Leftrightarrow}\)\(\begin{cases}x\ne1\\x\ne\pm1\end{cases}\)(bạn thông cảm chỗ này mình ko viết được ngoặc nhọn)

Vậy biểu thức A xác định khi \(x\ne\pm1\)

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

\(=\frac{2x^2+1}{x^2-1}=\frac{2x^2-2+3}{x^2-1}=\frac{2\left(x^2-1\right)+3}{x^2-1}=\frac{2\left(x^2-1\right)}{x^2-1}+\frac{3}{x^2-1}=2+\frac{3}{x^2-1}\)

c) A nguyên khi và chỉ khi  \(\frac{3}{x^2-1}\) nguyên 

<=>3 chia hết cho x²-1

<=>\(x^2-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

<=>\(x^2\in\left\{-2;0;2;4\right\}\)

Vì \(x^2\ge0\Rightarrow x^2\in\left\{0;2;4\right\}\)<=>\(x\in\left\{-2;0;\sqrt{2};2\right\}\)

Vì \(x\in Z\Rightarrow x\in\left\{-2;0;2\right\}\)

Vậy A nguyên khi \(x\in\left\{-2;0;2\right\}\)

a)A xác khi \(\hept{\begin{cases}x+1\ne0\\x^2-1\ne0\end{cases}\Rightarrow x\ne\left\{-1,1\right\}}\)

b) \(A=\frac{2x}{x+1}+\frac{1+2x}{\left(x-1\right)\left(x+1\right)}=\frac{2x\left(x-1\right)+1+2x}{\left(x-1\right)\left(x+1\right)}=\frac{2x^2+1}{x^2-1}=2+\frac{3}{\left(x^2\right)-1}\)

c)x^2-1=U(3)={-3,-1,1,3}

x^2={-2,0,2,4}

x={-2,0,2}

a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{x-2}\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

a: ĐKXĐ: 

b: 

a) x ≠ 2 và x  ≠  0

b) Rút gọn được Q = x + 1 2 x  

c) Thay x = 2017 (TMĐK) vào Q ta được Q = 1009 2017

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

\(\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\)

viết lại biểu thức 

a) \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}=\dfrac{\left(x-1-2\right)\left(x-1+2\right)}{\left(2x+1-x-2\right)\left(2x+1+x+2\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{3\left(x-1\right)\left(x+1\right)}\)  (1)

\(\Rightarrow\) ĐKXĐ: \(x\ne\pm1\)

b) \(\left(1\right)=\dfrac{x-3}{3x-3}\) (2)

c) Thay \(x=-3;x=1\) vào (2) ta có: \(\left\{{}\begin{matrix}B=\dfrac{-3-3}{3.\left(-3\right)-3}=\dfrac{1}{2}\\B=\dfrac{1-3}{3.1-3}=0\end{matrix}\right.\)

d) \(B=5\Rightarrow\dfrac{x-3}{3x-3}=5\Leftrightarrow x-3=15x-15\Leftrightarrow x=\dfrac{6}{7}\)