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a)\(\frac{\left(x-1\right)}{\sqrt{x}}\)
b) để P>0\(\Rightarrow\)\(\frac{\left(x-1\right)}{\sqrt{x}}>0\)
do \(\sqrt{x}>0\Rightarrow x-1>0\)
\(\Leftrightarrow x>1\)
c)P=\(\frac{8}{3}\)
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) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-x}-\frac{\sqrt{x}+2}{1-x}\)
\(\Leftrightarrow M=\frac{1}{1-\sqrt{x}}-\frac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow M=\frac{1+\sqrt{x}-\sqrt{x}-2}{1-x}\)
\(\Leftrightarrow M=\frac{-1}{1-x}\)
\(\Leftrightarrow M=\frac{1}{x-1}\)
b) Để M nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-1}\inℤ\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Mà \(x>0\)
Vậy để M nguyên \(\Leftrightarrow x=2\)
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne-1\\x\ne1\end{cases}}\)
Ta có \(P=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
b. Ta có \(P-\sqrt{x}=\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{x+2-x+\sqrt{x}}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Để \(P-\sqrt{x}\in Z\Rightarrow\sqrt{x}-1\inƯ\left(3\right)\Rightarrow\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
| \(\sqrt{x}-1\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(\sqrt{x}\) | -2 | 0 | 2 | 4 |
| x | 0 | 4 | 16 | |
| (l) | (n) | (n) | (n) |
Vậy \(x\in\left\{0;4;16\right\}\)thì \(P-\sqrt{x}\in Z\)
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)