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6 tháng 4 2018
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right]\) \(:\frac{\sqrt{x}+1-2}{x-1}\)
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right]:\frac{\sqrt{x}-1}{x-1}\)
\(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}:\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
6 tháng 4 2018
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\Leftrightarrow P=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}\)
\(\Leftrightarrow P=1-\frac{2}{\sqrt{x}+1}\)
để \(P\in Z\) \(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}+1\in\left\{\pm1;\pm2\right\}\)
+) \(\sqrt{x}+1=-1\Leftrightarrow\sqrt{x}=-2\) ( vô lí )
+) \(\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
+) \(\sqrt{x}+1=-2\Leftrightarrow\sqrt{x}=-3\) ( vô lí )
+) \(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\)
vậy để \(P\in Z\) thì \(x\in\left\{1;0\right\}\)
BA
30 tháng 7 2019
\(ĐKXĐ:\)
\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)
Vậy...................................................
BA
30 tháng 7 2019
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
10 tháng 8 2021
1.
\(a,Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(x+7\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)
\(b,Q\in Z\Leftrightarrow\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\in Z\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}\in Z\\ \Leftrightarrow-1-\dfrac{8}{\sqrt{x}-2}\in Z\)
Mà \(-1\in Z\Leftrightarrow\dfrac{8}{\sqrt{x}-2}\in Z\)
\(\Leftrightarrow8⋮\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;1;3;4;6;10\right\}\)
Mà \(x\in Z\) và \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1;4\right\}\\ \Leftrightarrow x\in\left\{0;1;4\right\}\)
Vậy \(x\in\left\{0;1;4\right\}\) thì \(Q\in Z\)
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