1A.
a) \(P=\dfrac{x-\sqrt[]{x}}{x-9}+\dfrac{1}{\sqrt[]{x}+3}-\dfrac{1}{\sqrt[]{x}-3}\left(x\ge0;x\ne9\right)\)
\(\Leftrightarrow P=\dfrac{x-\sqrt[]{x}+\sqrt[]{x}-3-\left(\sqrt[]{x}+3\right)}{x-9}\)
\(\Leftrightarrow P=\dfrac{x-\sqrt[]{x}+\sqrt[]{x}-3-\sqrt[]{x}-3}{x-9}\)
\(\Leftrightarrow P=\dfrac{x-\sqrt[]{x}-6}{x-9}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt[]{x}-3\right)\left(\sqrt[]{x}+2\right)}{\left(\sqrt[]{x}-3\right)\left(\sqrt[]{x}+3\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}+3}\)
b) -Với \(x=\sqrt[]{6+4\sqrt[]{2}}+\sqrt[]{6-4\sqrt[]{2}}\)
\(\Leftrightarrow x=\sqrt[]{4+2.2\sqrt[]{2}+2}+\sqrt[]{4-2.2\sqrt[]{2}+2}\)
\(\Leftrightarrow x=\sqrt[]{\left(2+\sqrt[]{2}\right)^2}+\sqrt[]{\left(2-\sqrt[]{2}\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt[]{2}\right|+\left|2-\sqrt[]{2}\right|\)
\(\Leftrightarrow x=2+\sqrt[]{2}+2-\sqrt[]{2}=4\)
Thay \(x=4\) vào \(P\) ta được
\(P=\dfrac{\sqrt[]{4}+2}{\sqrt[]{4}+3}=\dfrac{2+2}{2+3}=\dfrac{4}{5}\)
- Với \(x=\dfrac{1}{\sqrt[]{2}-1}-\dfrac{1}{\sqrt[]{2}+1}\)
\(\Leftrightarrow x=\dfrac{\sqrt[]{2}+1-\left(\sqrt[]{2}-1\right)}{\left(\sqrt[]{2}-1\right)\left(\sqrt[]{2}+1\right)}\)
\(\Leftrightarrow x=\dfrac{\sqrt[]{2}+1-\sqrt[]{2}+1}{2-1}\)
\(\Leftrightarrow x=2\)
Thay \(x=2\) vào \(P\) ta được
\(P=\dfrac{\sqrt[]{2}+2}{\sqrt[]{2}+3}=1-\dfrac{1}{\sqrt[]{2}+3}\)
1B:
a: \(Q=\dfrac{\sqrt{x}-2+7}{x-4}:\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+5}{x-4}\cdot\dfrac{\sqrt{x}-2}{1}=\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\)
b: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(=5+\sqrt{2}-4-\sqrt{2}=1\)
Khi x=1 thì \(Q=\dfrac{1+5}{1+3}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(x=\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(=\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3}+1=2\)
Khi x=2 thì \(Q=\dfrac{5+\sqrt{2}}{3+\sqrt{2}}=\dfrac{\left(5+\sqrt{2}\right)\left(3-\sqrt{2}\right)}{7}\)
