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: x<>4; x<>-4

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

c: Để A nguyên thì x+4-5 chia hết cho x+4

=>\(x+4\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-3;-5;1;-9\right\}\)

a) ĐK: \(x\ne4,x\ne2;x\ne-2\)

b) \(A=\dfrac{x^3}{x-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

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

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

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

\(A=x-1\)

c) \(A=0\) khi:

\(x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

d) A dương khi: \(A>0\)

\(x-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp với đk: 

\(x>1,x\ne4,x\ne2\)

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

1. ĐKXĐ: \(x\ne\pm1\)

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

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

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

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

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

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

Câu 6:

ĐKXĐ: \(x\ne-\dfrac{1}{3}\)

Để \(\dfrac{9x+4}{3x+1}\in Z\) thì \(9x+4⋮3x+1\)

=>\(9x+3+1⋮3x+1\)

=>\(1⋮3x+1\)

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

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

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

mà x nguyên

nên x=0

Câu 2:

a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)

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

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

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

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

a: ĐKXĐ: x<>-1

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

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

c: P=2

=>x^2-2x=2x+2

=>x^2-4x-2=0

=>\(x=2\pm\sqrt{6}\)

a, ĐKXĐ: \(a\ne1;a\ne-1\) 

Ta có:

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

\(\Rightarrow P=\frac{2a^2+a^2-a-a^2-a}{\left(a-1\right)\left(a+1\right)}=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}\)

\(\Rightarrow P=\frac{2a}{a+1}\) 

b. Để P có giá trị nguyên \(\Rightarrow2a⋮a+1\Rightarrow2\left(a+1\right)-2a⋮a+1\Rightarrow2a+2-2a⋮a+1\)

\(\Rightarrow2⋮a+1\) vì \(a\in Z\Rightarrow a+1\in\left\{-2;-1;1;2\right\}\Rightarrow a\in\left\{-3;-2;0;1\right\}\)

Vậy \(a\in\left\{-3;-2;0;1\right\}\)

Giúp Mk vs mai thi rồi

a khác 1 mà bạn