Bài 3 : với x >= 0 ; x khác 4
a, \(P=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{x\sqrt{x}+3x+2\sqrt{x}-2x+5\sqrt{x}-2+x-6\sqrt{x}+4}{x-4}=\dfrac{x\sqrt{x}+\sqrt{x}+2x+2}{x-4}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)+2\left(x+1\right)}{x-4}=\dfrac{x+1}{\sqrt{x}-2}\)
b, Ta có \(\sqrt{x}=\sqrt{9+4\sqrt{5}}=\sqrt{\left(2+\sqrt{5}\right)^2}=2+\sqrt{5}\)
\(\dfrac{2+\sqrt{5}+1}{2+\sqrt{5}-2}=\dfrac{3+\sqrt{5}}{\sqrt{5}}=\dfrac{3\sqrt{5}+5}{5}\)
Bài 4.
`a)`\(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(Q=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(Q=-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\)
\(Q=\dfrac{4\sqrt{x}}{x-4}\)
`b)`\(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)
\(Q=\dfrac{4\sqrt{\left(\sqrt{2}+1\right)^2}}{3+2\sqrt{2}-4}\)
\(Q=\dfrac{4\left(\sqrt{2}+1\right)}{2\sqrt{2}-1}\)
\(Q=\dfrac{4\sqrt{2}+4}{2\sqrt{2}-1}\)
