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a: \(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
\(=\dfrac{2\sqrt{x}+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}}{x-1}\)
\(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
\(\Rightarrow P=\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{x-1}\)
\(\Rightarrow P=\dfrac{\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}-1}\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{2+2\sqrt{2}}\)
\(\Rightarrow P=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}\)
\(\Rightarrow P=\dfrac{1}{2}\)
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\(a,P=\dfrac{3\sqrt{a}-3}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\left(a\ge0;a\ne1\right)\\ P=\dfrac{3\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{3\left(\sqrt{a}+1\right)}{\sqrt{a}}\\ b,a=4\Leftrightarrow\sqrt{a}=2\\ \Leftrightarrow P=\dfrac{3\left(2+1\right)}{2}=\dfrac{9}{2}\)
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1, \(\sqrt{\frac{-12}{x-5}}\) xác định khi \(\frac{-12}{x-5}\) \(\ge\) 0
→x-5<0→x<5
3. xác định khi x-2>0 →x>2
5.xác định khi \(\frac{4x-5}{x+2}\ge0\)và x\(\ne\)-2
→\(\left[\begin{array}{nghiempt}\hept{\begin{cases}4x-5< 0\\x-3< 0\end{array}\right.\\\hept{\begin{cases}4x-5\ge0\\x-3>0\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}x< \frac{5}{4}\\x< 3\end{array}\right.\\\hept{\begin{cases}x\ge\frac{5}{4}\\x>3\end{array}\right.\end{array}\right.}\)
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23.
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
Đáp án C
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{-2\sqrt{x}}=\dfrac{2x}{-2\sqrt{x}}=-\sqrt{x}\)
\(P=-\sqrt{x}=-\sqrt{4}=-2\left(đpcm\right)\)