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f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
a/ \(D\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\Rightarrow D=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b/\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)
\(=\sqrt[3]{5\sqrt{5}-3.5.1+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+3.5.1+3\sqrt{5}+1}\)
\(=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(\Rightarrow E=\sqrt{5}\)
c/
\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)
\(F^3=364+3F\sqrt[3]{182^2-33125}=364-3F\)
\(\Leftrightarrow F^3+3F-364=0\)
\(\Leftrightarrow\left(F-7\right)\left(F^2+7F+52\right)=0\)
\(\Rightarrow F=7\)
Bài 2:
a/ \(C=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)
\(=\sqrt{4}-1=2-1=1\)
Câu 1
a: \(=\dfrac{\sqrt{2}}{4}\)
b: \(=\sqrt{\dfrac{6}{100}}=\dfrac{\sqrt{6}}{10}\)
d: \(=\dfrac{7}{3\sqrt{3}}=\dfrac{7\sqrt{3}}{9}\)
Câu 2:
a: \(=\sqrt{\dfrac{1}{\sqrt{3}}}=\sqrt{\dfrac{\sqrt{3}}{3}}\)
b: \(=\sqrt{\dfrac{3\sqrt{3}+9}{6}}\)
c: \(=\dfrac{\sqrt{4}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=2\)
d: \(=\dfrac{5\sqrt{2}}{6}\)