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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 \(x\in\left\{1;0\right\}\)
16 tháng 10 2017
Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
16 tháng 11 2017
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/ \(\frac{x+\sqrt{x}-4}{x+2\sqrt{x}}=\frac{4+2\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}-4}{4+2\sqrt{3}+2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{4+2\sqrt{3}+\sqrt{3}+1-4}{4+2\sqrt{3}+2\sqrt{3}+2}=\frac{1+3\sqrt{3}}{6+4\sqrt{3}}\)
NC
18 tháng 8 2020
a) \(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)
NC
18 tháng 8 2020
b) \(P=\frac{7}{2}\)
\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}=\frac{7}{2}\)
\(\Rightarrow6\sqrt{x}+16=7\sqrt{x}+14\)
\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)
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
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\left(\sqrt{x}+3\right)}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4-x}\)
\(=\frac{3\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{3}{\left(2+\sqrt{x}\right)}\)
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