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Lời giải:
ĐKXĐ: $x\neq \pm 2$
\(A=\left[\frac{x}{(x-2)(x+2)}-\frac{2(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x+2)(x-2)}\right]:\frac{x^2-4+10-x^2}{x+2}\\ =\frac{x-2(x+2)+x-2}{(x-2)(x+2)}:\frac{6}{x+2}\\ =\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}\\ =\frac{-1}{x-2}=\frac{1}{2-x}\)
Để $A<0\Leftrightarrow \frac{1}{2-x}<0$
$\Leftrightarrow 2-x<0\Leftrightarrow x>2$
Kết hợp với ĐKXĐ suy ra $x>2$
b.
Với $x$ nguyên, để $A$ nguyên thì $1\vdots 2-x$
$\Rightarrow 2-x=1$ hoặc $2-x=-1$
$\Rightarrow x=1$ hoặc $x=3$
\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x}{x+2}\right)\)
\(=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x}{x+2}\right)\)
Đổi 10-x lại thành\(10-x^2\) nha, mk thiếu! sorry!
\(=\left(\frac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{6}\)
\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\) \(\left(ĐK:x\ne1;x\ne2\right)\)
\(=\frac{1-1+x}{1-x}:\frac{\left(1-x\right)\left(x+1\right)-\left(1-2x\right)}{1-x}\)
\(=\frac{x}{1-x}\cdot\frac{1-x}{1-x^2-1+2x}\)
\(=\frac{x}{-x^2+2x}\)
\(=\frac{x}{-x\left(x-2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
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
a) \(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)(ĐKXĐ: \(\hept{\begin{cases}x\ne\pm2\\x\ne3\end{cases}}\))\(=\left[\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}\right]:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)\(=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{2-x}{x-3}=\frac{4x}{x-3}\)
b) l\(x-5\)l\(=2\Leftrightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\left(n\right)\\x=3\left(l\right)\end{cases}\Rightarrow A=\frac{4.7}{7-3}=\frac{28}{4}=7}\)
c)
* Để A có giá trị là một số nguyên thì \(A=\frac{4x}{x-3}=\frac{4x-12+12}{x-3}=4+\frac{12}{x-3}\)là một số nguyên hay \(\frac{12}{x-3}\)là một số nguyên \(\Rightarrow x-3\inƯ\left(12\right)\Rightarrow S=\left(-9;-3;-1;0;1;4;5;6;7;9;15\right)\)(1)
* Để \(A=4+\frac{12}{x-3}< 4\Leftrightarrow\frac{12}{x-3}< 0\) thì \(x-3< 0\Leftrightarrow x< 3\)(2)
(1)(2) \(\Rightarrow S=\left(-9;-3;-1;0;1\right)\)
Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)