ĐKXĐ: x \(\) \(\)≠ {-2,0,2}

$\textbf{a)}$

$A=\left(\dfrac{x^2}{x(x-2)(x+2)}+\dfrac6{-3(x-2)}+\dfrac1{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)$

$=\left(\dfrac{x}{(x-2)(x+2)}-\dfrac2{x-2}+\dfrac1{x+2}\right):\left(\dfrac{(x-2)(x+2)+10-x^2}{x+2}\right)$

$=\left(\dfrac{x-2(x+2)+(x-2)}{(x-2)(x+2)}\right):\dfrac6{x+2}$

$=\dfrac{-6}{(x-2)(x+2)}\cdot\dfrac{x+2}{6}$

$=-\dfrac1{x-2}=\dfrac1{2-x}.$

$\textbf{b)}$

$|2x-1|=3$

$\Leftrightarrow\begin{cases}2x-1=3\\\text{hoặc}\\2x-1=-3\end{cases}$

$\Leftrightarrow\begin{cases}x=2\\\text{hoặc}\\x=-1.\end{cases}$

Do $x=2$ không thuộc ĐKXĐ nên $x=-1.$

$A=\dfrac1{2-(-1)}=\dfrac13.$

A)

a=42:3+6

a=20

B)

10=x:3+6

10-6=x:3

3=x:3

=> x=12

AI GIẢI HỘ MÌNH K CHO Ạ!!!

1)  a) Căn thức có nghĩa \(\Leftrightarrow4-2x\ge0\Leftrightarrow2x\le4\Leftrightarrow x\le2\)

b) Thay x = 2 vào biểu thức A, ta được: \(A=\sqrt{4-2.2}=\sqrt{0}=0\)

Thay x = 0 vào biểu thức A, ta được: \(A=\sqrt{4-2.0}=\sqrt{4}=2\)

Thay x = 1 vào biểu thức A, ta được: \(A=\sqrt{4-2.1}=\sqrt{2}\)

Thay x = -6 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-6\right)}=\sqrt{16}=4\)

Thay x = -10 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-10\right)}=\sqrt{24}=2\sqrt{6}\)

c) \(A=0\Leftrightarrow\sqrt{4-2x}=0\Leftrightarrow4-2x=0\Leftrightarrow x=2\)

\(A=5\Leftrightarrow\sqrt{4-2x}=5\Leftrightarrow4-2x=25\Leftrightarrow x=\frac{-21}{2}\)

\(A=10\Leftrightarrow\sqrt{4-2x}=10\Leftrightarrow4-2x=100\Leftrightarrow x=-48\)

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 \(x=\left\{2;0;4;-2\right\}\)

1: ĐKXĐ: x∉{1;-1}

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

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

\(=\frac{x^2+2x+1-\left(x^2+2x-3\right)}{x-1}\cdot\frac12=\frac{4}{2\left(x-1\right)}=\frac{2}{x-1}\)

3: THay x=5 vào A, ta được:

\(A=\frac{2}{5-1}=\frac24=\frac12\)

4: Để A là số nguyên thì 2⋮x-1

=>x-1∈{1;-1;2;-2}

=>x∈{2;0;3;-1}

Kết hợp ĐKXĐ, ta được: x∈{0;2;3}

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

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

a: 

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

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

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

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

Khi x=2005 thì \(A=\dfrac{-1}{2\cdot\left(2005-1\right)}=-\dfrac{1}{4008}\)

Vì x=1 không thỏa mãn ĐKXĐ

nên khi x=1 thì A không có giá trị

c: Để A=-1002 thì \(\dfrac{-1}{2\left(x-1\right)}=-1002\)

=>\(2\left(x-1\right)=\dfrac{1}{1002}\)

=>\(x-1=\dfrac{1}{2004}\)

=>\(x=\dfrac{1}{2004}+1=\dfrac{2005}{2004}\left(nhận\right)\)