Trục căn thức ở mẫu
a)\(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
b)\(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
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27 tháng 9 2016
a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)
c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)
d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)
19 tháng 6 2019
a) \(\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b) \(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}=\sqrt{5}+\sqrt{2}\)
c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3}{5-3}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}\)
d) \(\frac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\frac{1}{3\sqrt{2}+2\sqrt{2}-2\sqrt{2}}=\frac{1}{3\sqrt{2}}=\frac{\sqrt{2}}{3\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{2}}{6}\)
NQ
5 tháng 8 2019
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..