1B:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
\(=\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1-\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
=1
Thay x=1 vào Q, ta được:
\(Q=\dfrac{1+5}{1+2}=\dfrac{6}{3}=2\)
Ta có: \(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
Thay x=2 vào Q,ta được:\(Q=\dfrac{5+\sqrt{2}}{2+\sqrt{2}}=\dfrac{\left(5+\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}\)
\(=\dfrac{10-5\sqrt{2}+2\sqrt{2}-2}{2}=\dfrac{8-3\sqrt{2}}{2}\)
\(a.\\ P=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(b.\\ b.1.\\ x=\sqrt{4+4\sqrt{2}+2}+\sqrt{4-4\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|+\left|2-\sqrt{2}\right|=2+\sqrt{2}+2-\sqrt{2}=4\)
\(b.2.\\ x=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=2\)
