a) Ta có: \(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
b) Thay \(x=9+4\sqrt{5}\) vào P, ta được:
\(P=\dfrac{\left(9+4\sqrt{5}\right)\left(\sqrt{5}+2\right)+2\left(9+4\sqrt{5}+2\right)}{9+4\sqrt{5}-4}\)
\(=\dfrac{9\sqrt{5}+18+20+8\sqrt{5}+22+8\sqrt{5}}{5+4\sqrt{5}}\)
\(=\dfrac{25\sqrt{5}+60}{\sqrt{5}\left(\sqrt{5}+4\right)}\)
\(=\dfrac{\sqrt{5}\left(25+12\sqrt{5}\right)}{\sqrt{5}\left(\sqrt{5}+4\right)}\)
\(=\dfrac{\left(25+12\sqrt{5}\right)\left(\sqrt{5}-4\right)}{-11}\)
\(=\dfrac{25\sqrt{5}-100+60-48\sqrt{5}}{-11}\)
\(=\dfrac{-23\sqrt{5}-40}{-11}=\dfrac{23\sqrt{5}+40}{11}\)
a) \(P=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{x\sqrt{x}+3x+2\sqrt{x}-\left(2x-5\sqrt{x}+2\right)+x-6\sqrt{x}+4}{x-4}\\ =\dfrac{x\sqrt{x}+2x+\sqrt{x}+2}{x-4}=\dfrac{\sqrt{x}\left(x+1\right)+2\left(x+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x+1}{\sqrt{x}-2}\)
b) \(x=9+4\sqrt{5}\\ \Rightarrow P=\dfrac{10+4\sqrt{5}}{\sqrt{9+4\sqrt{5}-2}}=\dfrac{10+4\sqrt{5}}{\sqrt{\left(\sqrt{5}+2\right)^2-2}}=\dfrac{10-4\sqrt{5}}{\sqrt{5}+2-2}\\ =\dfrac{10+4\sqrt{5}}{\sqrt{5}}=2\sqrt{5}+4\)
