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\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{x-2\left(x+2\right)+1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{-6}{x^2-4}\right):\left(\dfrac{6}{x+2}\right)\)
\(\Rightarrow A=-\dfrac{6}{x^2-4}.\dfrac{x+2}{6}=-\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)6}=-\dfrac{1}{x-2}\)
để A<0 thì :
\(\left\{{}\begin{matrix}x-2\ne0\\x-2\notin Z-\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{3;4;5;6;7;8;9;....n\right\}\)
( Z- là tập hợp số nguyên âm )
Để A có giá trị nguyên thì :
\(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
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a) ĐKXĐ: x∉{2;-2}
b) Ta có: \(A=\dfrac{x}{x-2}+\dfrac{2-x}{x+2}+\dfrac{12-10x}{x^2-4}\)
\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-x^2+4x-4+12-10x}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4x+8}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4}{x+2}\)
c) Để \(A=\dfrac{2}{3}\) thì \(\dfrac{-4}{x+2}=\dfrac{2}{3}\)
\(\Leftrightarrow x+2=\dfrac{-4\cdot3}{2}=-\dfrac{12}{2}=-6\)
hay x=-6-2=-8(nhận)
Vậy: Để \(A=\dfrac{2}{3}\) thì x=-8
d) Để A nguyên thì \(-4⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(-4\right)\)
\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{-1;-3;0;-4;2;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)
b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Khi x=-1/2 thì B=2/5
c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a, đk : x khác -2 ; 2
\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)
b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)
Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)
c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)
2-x | 1 | -1 |
x | 1 | 3 |
![](https://rs.olm.vn/images/avt/0.png?1311)
Để B là số nguyên thì \(4⋮x^2-2x+2\)
=>\(x^2-2x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\left(x-1\right)^2+1\in\left\{1;-1;2;-2;4;-4\right\}\)
mà \(\left(x-1\right)^2+1>=1\forall x\)
nên \(\left(x-1\right)^2+1\in\left\{1;2;4\right\}\)
=>\(\left(x-1\right)^2\in\left\{0;1;3\right\}\)
=>\(x-1\in\left\{0;1;-1;\sqrt{3};-\sqrt{3}\right\}\)
=>\(x\in\left\{1;2;0;\sqrt{3}+1;1-\sqrt{3}\right\}\)
mà x nguyên
nên \(x\in\left\{1;2;0\right\}\)