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![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)
\(=\left[\frac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}-\frac{\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\right]\)
Phương trình tương đương :
\(=\frac{2x^2-2x}{x^2-x}:\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2:\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\frac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(ĐKXĐ:x\ge0;x\ne1\)
= \(\frac{x+1+\sqrt{x}}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\frac{x+2}{\sqrt{x}-1}\)
B ) Ta có :
\(Q=P-\sqrt{x}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}-\sqrt{x}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Đế Q nhận giá trị nguyên thì \(1+\frac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\in Z\left(vì1\in Z\right)\)
\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)\)
Ta có bảng sau :
\(\sqrt{x}-1\) | 3 | -3 | 1 | -1 |
\(\sqrt{x}\) | 4 | -2 | 2 | 0 |
\(x\) | 16(t/m) | 4(t/m) | 0(t/m) |
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì \(x\in\left\{16;4;0\right\}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right).\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x+2}{\sqrt{x}+1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x+\sqrt{x}-x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
Để P âm \(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Mà \(\sqrt{x}+2>0\forall x\Rightarrow\sqrt{x}-1< 0\Rightarrow x< 1\)
Để \(P\in Z\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\in Z\)
\(\Rightarrow1-\frac{3}{\sqrt{x}+2}\in Z\Rightarrow\frac{3}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ_3\)
Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\Rightarrow x=1\)
Mà để \(P\in Z^-\Rightarrow\hept{\begin{cases}x< 1\\x=1\end{cases}}\)\(\Rightarrow x\in\varnothing\)
Vậy không có giá trị nào của x để P nguyên âm