B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)

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

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

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)

b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)

\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))

\(\Leftrightarrow x>-1\).

-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

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

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị

\(a,A=\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{1}{\sqrt{x}-1}\\ b,A< 0\Leftrightarrow\dfrac{1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\left(1>0\right)\\ \Leftrightarrow x< 1\\ c,A\in Z\Leftrightarrow1⋮\sqrt{x}-1\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(1\right)\left\{-1;1\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{0;4\right\}\)

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

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

b) \(A=\dfrac{1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Kết hợp đk: 

\(\Rightarrow0\le x< 1\)

c) \(A=\dfrac{1}{\sqrt{x}-1}\in Z\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2\right\}\)

\(\Rightarrow x\in\left\{0;4\right\}\)

Giúp mk vs ,mk cần gấp

a) Ta có: \(A=\left(\dfrac{2}{x+2}-\dfrac{1}{x-3}+\dfrac{5-x}{x^2-x-6}\right)\cdot\left(x-\dfrac{6}{x-1}\right)\)

\(=\left(\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}+\dfrac{5-x}{\left(x-3\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-1\right)-6}{x-1}\)

\(=\dfrac{2x-6-x-2+5-x}{\left(x+2\right)\left(x-3\right)}\cdot\dfrac{x^2-x-6}{x-1}\)

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

a) \(-ĐKXĐ:x\ne\pm2;1\)

Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

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

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

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

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

b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)

mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)

mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để P>0 thì x>1

\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\)     \(\left(ĐK:x\ne1;x\ne2\right)\)

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

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

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

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

b) Để A=\(\frac{1}{2}\) \(\Leftrightarrow\)\(\frac{1}{2-x}=\frac{1}{2}\)

                   \(\Leftrightarrow2-x=2\)

                   \(\Leftrightarrow-x=0\Leftrightarrow x=0\)

c) Để A>1 \(\Leftrightarrow\)\(\frac{1}{2-x}>1\)

                 \(\Leftrightarrow\)\(\frac{1}{2-x}-1>0\) 

                 \(\Leftrightarrow\)\(\frac{1-2+x}{2-x}>0\)

                 \(\Leftrightarrow\)\(\frac{x-1}{2-x}>0\)

\(\Leftrightarrow\begin{cases}x-1>0\\2-x>0\end{cases}\) hoặc \(\begin{cases}x-1< 0\\2-x< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>1\\x< 2\end{cases}\) hoặc \(\begin{cases}x< 1\\x>2\end{cases}\)(vô nghiệm)

\(\Leftrightarrow1< x< 2\)

Vậy \(1< x< 2\) thì A<1