`a.`\(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\);\(x\ge0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
\(Q=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\)
\(Q=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
`b.`
`@`\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(x=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(x=\left|5+\sqrt{2}\right|-\left|4+\sqrt{2}\right|\)
\(x=5+\sqrt{2}-4-\sqrt{2}\)
\(x=1\) thế vào `Q`
\(Q=\dfrac{\sqrt{1}+5}{\sqrt{1}+2}=\dfrac{6}{3}=2\)
`@`\(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(x=\dfrac{2}{\sqrt{4-2\sqrt{3}}}-\dfrac{2}{\sqrt{4+2\sqrt{3}}}\)
\(x=\dfrac{2}{\sqrt{\left(\sqrt{3}-1\right)^2}}-\dfrac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(x=\dfrac{2}{\left|\sqrt{3}-1\right|}-\dfrac{2}{\left|\sqrt{3}+1\right|}\)
\(x=\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(x=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(x=\dfrac{4}{2}=2\) thế vào `Q`
\(Q=\dfrac{\sqrt{2}+5}{\sqrt{2}+2}=\dfrac{8-3\sqrt{2}}{2}\)
\(Q=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{1}{\sqrt{x}-2}\right)\)
\(=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b.
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(=5+\sqrt{2}-\left(4+\sqrt{2}\right)=1\)
\(\Rightarrow Q=\dfrac{1+5}{1+2}=2\)
\(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\dfrac{2}{\sqrt{4-2\sqrt{3}}}-\dfrac{2}{\sqrt{4+2\sqrt{3}}}=\dfrac{2}{\sqrt{\left(\sqrt{3}-1\right)^2}}-\dfrac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}=\dfrac{2\left(\sqrt{3}+1-\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{2.2}{2}=2\)
\(\Rightarrow Q=\dfrac{\sqrt{2}+5}{\sqrt{2}+2}=\dfrac{8-3\sqrt{2}}{2}\)
a. \(Q=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
\(=\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
Vậy : \(Q=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b1. Ta có : \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}+4\right)^2}\)
\(=\left|\sqrt{2}+5\right|-\left|\sqrt{2}+4\right|=1\left(tmđk\right)\)
Thay giá trị x trên vào Q \(\Rightarrow Q=\dfrac{\sqrt{1}+5}{\sqrt{1}+2}=2\)
Vậy : Với \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\) thì Q = 2.
b2.
Ta có : \(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2\left(2+\sqrt{3}\right)}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\dfrac{\sqrt{2\left(2-\sqrt{3}\right)}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}-\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{1}}-\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{1}}\)
\(=\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|=2\left(tmđk\right)\)
Thay giá trị x trên vào Q \(\Rightarrow Q=\dfrac{\sqrt{2}+5}{\sqrt{2}+2}=\dfrac{8-3\sqrt{2}}{2}\)
Vậy : Với \(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\) thì \(Q=\dfrac{8-3\sqrt{2}}{2}\)