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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) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}-\frac{1-\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\right):\left(\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{1-\sqrt{x}}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x-\sqrt{x}+1}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1+\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}.\frac{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)
\(\Rightarrow P=\frac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}=\frac{6-3\sqrt{3}}{2-\sqrt{3}}=3\)
Câu c hơi nghi ngờ cái đề, cấp 2 làm sao giải được BPT bậc 3 kiểu này?