a) ĐKXĐ: a²-1 ≠0 ⇔ (a-1)(a+1)≠0 ⇔\(\left[{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)

b) A=\(\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\) , a≠1, -1

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

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

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

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

vậy A =\(\dfrac{2a}{a-1}\) với a≠1,-1.

c) Có:A= \(\dfrac{2a}{a-1}\) = \(\dfrac{2a-2+2}{a-1}=\dfrac{2\left(a-1\right)+2}{a-1}=2+\dfrac{2}{a-1}\)

Để a∈Z thì a-1 ∈ Z ⇒ (a-1) ∈ Ư(2) =\(\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

Vậy để biểu thức A có giá trị nguyên thì a∈\(\left\{2;0;3\right\}\)

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

b) Ta có: \(A=\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\)

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

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

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

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

\(=\dfrac{2a}{a-1}\)

c) Để A nguyên thì \(2a⋮a-1\)

\(\Leftrightarrow2a-2+2⋮a-1\)

mà \(2a-2⋮a-1\)

nên \(2⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(2\right)\)

\(\Leftrightarrow a-1\in\left\{1;-1;2;-2\right\}\)

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

Kết hợp ĐKXĐ, ta được: \(a\in\left\{0;2;3\right\}\)

Vậy: Để A nguyên thì \(a\in\left\{0;2;3\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)

\(a,A=\dfrac{x^2-3x+2+x^2+3x+2-x^2+2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+2x}{\left(x+2\right)\left(x-2\right)}\\ A=\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x}{x-2}\\ b,A=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\in Z\\ \Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Rightarrow x\in\left\{0;1;3;4\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

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

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

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

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

b: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\inƯ\left(2\right)\)

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

=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)

=>\(x\in\left\{9;1;16;0\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;16\right\}\)

c: A<0

=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

Kết hợp ĐKXĐ, ta được: 0<x<4 và x<>1

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

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

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

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

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

b: \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

c: \(C=-\dfrac{1}{2}\)

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

=>2x+2=-2

=>2x=-4

=>x=-2(nhận)

d: Để C là số nguyên thì \(2x+2\inƯ\left(1\right)\)

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

=>\(2x\in\left\{-1;-3\right\}\)

=>\(x\in\left\{-\dfrac{1}{2};-\dfrac{3}{2}\right\}\)