1,
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{\sqrt{x^3}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\\ =\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\\ =\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\\ =\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\\ =\frac{x+2\sqrt{x}+1}{\sqrt{x}}\left(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\right)\)
Ta có:
\(P=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{9}{2}\\ \Leftrightarrow x+2\sqrt{x}+1=\frac{9}{2}\sqrt{x}\\ \Leftrightarrow2x+4\sqrt{x}+2=9\sqrt{x}\\ \Leftrightarrow2x-5\sqrt{x}+2=0\\ \Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
2, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có:
\(P=\left(\frac{\sqrt{x^3}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}-1}\right)\\ =\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-1}{\sqrt{x}-1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\right)\\ =\left(\frac{x-\sqrt{x}+1-x+1}{\sqrt{x}-1}\right):\frac{x}{\sqrt{x}-1}\\ =\frac{2-\sqrt{x}}{\sqrt{x}-1}\cdot\frac{\sqrt{x}-1}{x}\\ =\frac{2-\sqrt{x}}{x}\)
Ta có:
\(P=\frac{2-\sqrt{x}}{x}=3\\ \Leftrightarrow2-\sqrt{x}=3x\\ \Leftrightarrow3x+\sqrt{x}-2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=\frac{2}{3}\end{matrix}\right.\\ \Leftrightarrow x=\frac{4}{9}\)
