a)
Ta có: \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{x-1}-1\right)\)
\(=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}}{x-1}-\frac{x-1}{x-1}\right)\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}+1}{\sqrt{x}-x+1}\)
b) ĐKXĐ: \(x\ne1\); \(x\ge0\)
Để \(P=1\frac{2}{3}\) thì \(\frac{2\sqrt{x}+1}{\sqrt{x}-x+1}=1\frac{2}{3}\)
\(\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}-x+1}=\frac{5}{3}\)
\(\Leftrightarrow3\left(2\sqrt{x}+1\right)=5\left(\sqrt{x}-x+1\right)\)
\(\Leftrightarrow6\sqrt{x}+3=5\sqrt{x}-5x+5\)
\(\Leftrightarrow6\sqrt{x}+3-5\sqrt{x}+5x-5=0\)
\(\Leftrightarrow5x+\sqrt{x}-2=0\)
Đến đây bạn tự làm tiếp nhé
