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a. \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
b. \(\dfrac{26}{5-2\sqrt{3}}=\dfrac{26\left(5+2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}=\dfrac{26\left(5+2\sqrt{3}\right)}{13}=2\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
c. \(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}=\dfrac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\dfrac{3\sqrt{10}}{6}=\dfrac{\sqrt{10}}{2}\)
d. \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\dfrac{23\sqrt{6}}{46}=\dfrac{\sqrt{6}}{2}\)
Lời giải:
a) \(\frac{1}{1-\sqrt[3]{5}}=\frac{1+\sqrt[3]{5}+\sqrt[3]{5^2}}{(1-\sqrt[3]{5})(1+\sqrt[3]{5}+\sqrt[3]{25})}\) \(=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{1^3-5}=\frac{1+\sqrt[3]{5}+\sqrt[3]{25}}{-4}\)
b)
\(\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}}=\frac{\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2}}{(\sqrt[3]{2}+\sqrt[3]{3})(\sqrt[3]{2^2}-\sqrt[3]{6}+\sqrt[3]{3^2})}\) \(=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{2+3}=\frac{\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}}{5}\)
c)
\(\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{2^2}+\sqrt[3]{2}+1)}=\frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)
a=1/(√3+√2+1)=(√3-(√2+1)/[3-(√2+1)^2]
=(√3-√2-1)/(3-(3+2√2)
=(√3-√2-1)/(-2√2)
=-(√6-2-√2)/4
=(2+√2-√6)/4
a) \(\dfrac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}\dfrac{\sqrt{2}+2+\sqrt{6}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}+3-3}=\dfrac{\sqrt{2}+2+\sqrt{6}}{2\sqrt{2}}=\dfrac{1+\sqrt{2}+\sqrt{3}}{2}\)
b) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}+5-5}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{2\sqrt{6}\cdot\sqrt{6}}=\dfrac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}\)
a: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{7+2\sqrt{10}-3}=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{4+2\sqrt{10}}\)
\(=\dfrac{-\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(4-2\sqrt{10}\right)}{24}\)
b: \(=\dfrac{2+\sqrt{3}+\sqrt{5}}{4-8+2\sqrt{15}}=\dfrac{2+\sqrt{3}+\sqrt{5}}{2\sqrt{15}-4}\)
\(=\dfrac{\left(2+\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+4\right)}{44}\)
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}\right)^2-2}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{5+3-2-2\sqrt{15}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{6-2\sqrt{15}}=\dfrac{\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{15}\right)}{\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)2}=\dfrac{3\sqrt{5}-3\sqrt{3}-3\sqrt{2}+5\sqrt{3}-3\sqrt{5}-\sqrt{30}}{\left(9-15\right).2}=\dfrac{2\sqrt{3}-3\sqrt{2}-\sqrt{30}}{-12}\)b. \(\dfrac{1}{2-\sqrt{3}-\sqrt{5}}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}+\sqrt{5}\right)}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}\right)^2-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{4-4\sqrt{3}+3-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{2-4\sqrt{3}}=\dfrac{\left(2-\sqrt{3}+\sqrt{5}\right)\left(1+2\sqrt{3}\right)}{2\left(1-2\sqrt{3}\right)\left(1+2\sqrt{3}\right)}=\dfrac{2+4\sqrt{3}-\sqrt{3}-6+\sqrt{5}+2\sqrt{15}}{2.\left(1-12\right)}=\dfrac{3\sqrt{3}+\sqrt{5}+2\sqrt{15}-4}{-22}\)
\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}+1-1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}+1-1}\)
\(=\dfrac{\sqrt{3}\left[\left(\sqrt{\sqrt{3}+1}+1\right)-\left(\sqrt{\sqrt{3}+1}-1\right)\right]}{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\)
\(=2\)