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1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(x-1\right)}=\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x-\sqrt{x}}{x-\sqrt{x}}< 0\)\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x-\sqrt{x}}< 0\)
\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))
\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)
\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)
a) \(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)ư
=\(\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x +2\sqrt{x}-7-x+\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)
=\(\dfrac{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
b)ta có : \(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)
để P nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\Leftrightarrow\sqrt{x}-3\inƯ\left(\pm1,\pm2\right)\)
\(\Rightarrow\sqrt{x}-3=1\Leftrightarrow x=16\left(TM\right)\)
\(\sqrt{x}-3=-1\Leftrightarrow x=4\left(KTM\right)\)
\(\sqrt{x}-3=2\Leftrightarrow x=25\left(TM\right)\)
\(\sqrt{x}-3=-2\Leftrightarrow x=1\left(KTM\right)\)
vậy x\(\in\left\{16,25\right\}\)
a, Mk làm hơi tắt chút bạn thông cảm nha . mk vội ý mà 
\(A=\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)
\(A=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu c : \(A\in Z\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}\in Z\Leftrightarrow1-\dfrac{1}{\sqrt{x}}\in Z\)
Để : \(1-\dfrac{1}{\sqrt{x}}\in Z\) thì \(\sqrt{x}\inƯ\left(1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=1\)
a: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-x}{x-1}\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{-x+\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(-x+\sqrt{x}+1\right)}\)
b: Để A là số nguyên thì \(\left(\sqrt{x}-1\right)^2⋮\left(\sqrt{x}+1\right)\left(-x+\sqrt{x}+1\right)\)
=>x=0
a:
Sửa đề: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}+2}{x\sqrt{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{x}{x-1}\right)\)
\(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-x}{x-1}\)
\(=\dfrac{x-1-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{-x+\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{-x+\sqrt{x}+1}=\dfrac{-\sqrt{x}+3}{x-\sqrt{x}-1}\)
b: Để A là số nguyên thì \(\sqrt{x}\left(-\sqrt{x}+3\right)⋮x-\sqrt{x}-1\)
=>\(-x+3\sqrt{x}⋮x-\sqrt{x}-1\)
=>\(-x+\sqrt{x}+1+2\sqrt{x}-1⋮x-\sqrt{x}-1\)
=>\(x=0\)
a) P = \(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
= \(\left(\dfrac{-\sqrt{x}+\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\left(\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
= \(\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\) = \(\dfrac{1}{\sqrt{x}+1}.\dfrac{\sqrt{x}-2}{1}\) = \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a/ \(\frac{x-2}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}+2}\)
\(=\frac{x-2}{x+2\sqrt{x}}-\frac{\sqrt{x}+2}{x+2\sqrt{x}}+\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(=\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}\)
b/ \(\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}=\frac{4+2\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}-4}{4+2\sqrt{3}+2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{4+2\sqrt{3}+\sqrt{3}+1-4}{4+2\sqrt{3}+2\sqrt{3}+2}=\frac{1+3\sqrt{3}}{6+4\sqrt{3}}\)
Để \(A\in Z\) thì \(\sqrt{x}+3\) phải chia hết cho \(\sqrt{x}-2\).
\(\Rightarrow\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)⋮\left(\sqrt{x}-2\right)\)
\(\Rightarrow5⋮\left(\sqrt{x}-2\right)\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(5\right)\)
\(\Rightarrow\sqrt{x}-2=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\sqrt{x}=\left\{3;7;1;-3\right\}\)
\(\Rightarrow x=\left\{9;49;1\right\}\)