Điều kiện xác định của \(\sqrt{2-x}+\sqrt{x-2}\)là x=?
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK:
\(\left\{{}\begin{matrix}x-1\ge0\\2-x>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x< 2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4x+2\ge0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge2+\sqrt{2}\\x\le2-\sqrt{2}\end{matrix}\right.\\x\ge2\end{matrix}\right.\)
\(\Rightarrow x\ge2+\sqrt{2}\)
\(x^2-4x+2\ge0\Leftrightarrow x^2-4x+4\ge2\)
\(\Leftrightarrow\left(x-2\right)^2\ge2\)
\(\Leftrightarrow\left|x-2\right|\ge\sqrt{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge\sqrt{2}\\x-2\le-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2+\sqrt{2}\\x\le2-\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{3x+2\sqrt{x}-5}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{1-\sqrt{x}}\)
\(=\dfrac{3x+2\sqrt{x}-5+\sqrt{x}-1+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: (x-1)(x-3)>=0
=>x>=3 hoặc x<=1
b: ĐKXĐ: (x-4)(x-3)>=0
=>x>=4 hoặc x<=3
c: ĐKXĐ: (x-5)(x-4)>=0
=>x>=5 hoặc x<=4
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. ĐKXĐ:
\(\left\{\begin{matrix} x-1\geq 0\\ 2\geq \sqrt{x-1}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 4\geq x-1\end{matrix}\right. \Leftrightarrow 5\geq x\geq 1\)
b. ĐKXĐ:
\(\left\{\begin{matrix} x\geq 0\\ 3\geq \sqrt{x}\end{matrix}\right.\Leftrightarrow 0\leq x\leq 9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: `{(5x-1>=0),(x+2>=0),(7-x>=0):}`
`<=>{(x>=1/5),(x>=-2),(x<=7):}`
`<=>1/5 <=x<=7`
`ĐKXĐ: {(5x - 1 >= 0),(x+2 >=0),(7-x >=0):}`
`<=> {(x >= 1/5),(x>= -2),(x <=7):}`
`<=> 1/5 <= x <= 7`
x = 2