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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 \(x\in\left\{16;4;0\right\}\)
\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(=1-\left(\frac{2\left(1-2\sqrt{x}\right)+5\sqrt{x}-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}\right):\frac{\sqrt{x}-1}{\left(1+2\sqrt{x}\right)^2}\)
\(=1-\frac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}.\frac{\left(1+2\sqrt{x}\right)^2}{\sqrt{x}-1}=1-\frac{1+2\sqrt{x}}{1-2\sqrt{x}}=2-\frac{2}{1-2\sqrt{x}}\)
để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(K=\left(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\left(\frac{x+3\sqrt{x}+\sqrt{x}-1-x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\left(\sqrt{x}-1\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+2}\)
\(K=\frac{2\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=2-\frac{5}{\sqrt{x}+2}\)
\(\Rightarrow5⋮\left(\sqrt{x}+2\right)\) mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=5\Rightarrow x=9\)
c/ \(\frac{5}{\sqrt{x}+2}>0\Rightarrow2-\frac{5}{\sqrt{x}+2}< 2\Rightarrow K< 2\)
d/ \(\sqrt{x}+2\ge2\Rightarrow2-\frac{5}{\sqrt{x}+2}\ge2-\frac{5}{2}=-\frac{1}{2}\)
\(\Rightarrow K_{min}=-\frac{1}{2}\) khi \(x=0\)
e/ \(K=\frac{2\sqrt{x}-1}{\sqrt{x}+2}\Leftrightarrow K\sqrt{x}+2K=2\sqrt{x}-1\)
\(\Leftrightarrow\left(K-2\right)\sqrt{x}=-2K-1\Rightarrow\sqrt{x}=\frac{2K+1}{2-K}\)
Mà \(\sqrt{x}\ge0\Rightarrow\frac{2K+1}{2-K}\ge0\Rightarrow-\frac{1}{2}\le K< 2\)
\(\Rightarrow K=\left\{0;1\right\}\)
- Với \(K=0\Rightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+2}=0\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
- Với \(K=1\Rightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+2}=1\Rightarrow2\sqrt{x}-1=\sqrt{x}+2\Rightarrow x=9\)
cảm ơn ạ