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a: \(P=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x}\)
b: Để P<0 thì \(\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)< 0\)
=>1<x<4
a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)
\(M=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+2}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)
Ta có: P = \frac{4\sqrt{x}}{8x} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} - 2} : \frac{\sqrt{x} + 2}{x - 4} \cdot \frac{\sqrt{x} - 2}{\sqrt{x} + 2} = \frac{4\sqrt{x}(\sqrt{x} + 2)}{(8x)(\sqrt{x} - 2)} : \frac{x - 4}{x - 4} = \frac{4(\sqrt{x} + 2)}{8(\sqrt{x} - 2)} = \frac{1}{\sqrt{x} - 2} 2) Tìm các giá trị của x để P = -4: Ta có: P = -4 \Rightarrow \frac{1}{\sqrt{x} - 2} = -4 \Rightarrow \sqrt{x} - 2 = -\frac{1}{4} \Rightarrow \sqrt{x} = \frac{7}{4} \Rightarrow x = \left(\frac{7}{4}\right)^2 = \frac{49}{16} Vậy x = 49/16 là giá trị cần tìm.
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a) \(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{x-1}\right)\)
\(P=\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\div\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\div\frac{x-\sqrt{x}+\sqrt{x}-4}{x-1}\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\times\frac{x-1}{x-4}\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\)
\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-4}\)
\(P=\frac{x-3\sqrt{x}+2}{x-4}\)
b) Để P < 0
=> \(\frac{x-3\sqrt{x}+2}{x-4}< 0\)
Xét hai trường hợp
I) \(\hept{\begin{cases}x-3\sqrt{x}+2>0\\x-4< 0\end{cases}}\)
+) \(x-3\sqrt{x}+2>0\)
<=> ( √x - 1 )( √x - 2 ) > 0
1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>4\end{cases}}\Leftrightarrow x>4\)(1)
2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 1\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 4\end{cases}}\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ : \(0\le x< 1\)(2)
+) x - 4 < 0 <=> x < 4 (3)
Từ (1), (2) và (3) => \(0\le x< 1\)
II) \(\hept{\begin{cases}x-3\sqrt{x}+2< 0\\x-4>0\end{cases}}\)
+) \(x-3\sqrt{x}+2< 0\)
<=> ( √x - 1 )( √x - 2 ) < 0
1. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 1\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>4\end{cases}}\)( loại )
2. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 4\end{cases}}\Leftrightarrow1< x< 4\)(1)
+) x - 4 > 0 <=> x > 4 (2)
Từ (1) và (2) => Không có giá trị của x thỏa mãn
Vậy với \(0\le x< 1\)thì P < 0