a) Ta có: \(P=\left(\frac{x-3\sqrt{x}}{x-6\sqrt{x}+9}-\frac{2\sqrt{x}-1}{x-3\sqrt{x}}\right)\cdot\frac{x-9}{\sqrt{x}+3}\)
\(=\left(\frac{\sqrt{x}\left(x-3\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-3\right)^2}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\right)\cdot\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\frac{x\sqrt{x}-3x-\left(2x-6\sqrt{x}-\sqrt{x}+3\right)}{\sqrt{x}\cdot\left(\sqrt{x}-3\right)^2}\cdot\left(\sqrt{x}-3\right)\)
\(=\frac{x\sqrt{x}-3x-2x+6\sqrt{x}+\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-5x+7\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-1-5x+5\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1-5\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x-4\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}-3\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\cdot\left[\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)\right]}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}-3\right)}{\sqrt{x}\cdot\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
b)ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)
Vì x=0.25 thỏa mãn ĐKXĐ nên Thay x=0.25 vào biểu thức \(P=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\), ta được:
\(P=\frac{\left(\sqrt{0.25}-1\right)^2}{\sqrt{0.25}}=\frac{\left(0.5-1\right)^2}{0.5}=\frac{\left(-0.5\right)^2}{0.5}=\frac{0.25}{0.5}=0.5\)
