\(\sqrt{18}-\frac{1}{3}\sqrt{72}-\sqrt{8}+\frac{2-3\sqrt{2}}{3-\sqrt{2}}\)
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11 tháng 9 2016
\(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{23}\)
31 tháng 8 2021
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
M
0
K
3
3 tháng 7 2020
\(\frac{5}{2+\sqrt{3}}=\frac{5\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{5\left(2-\sqrt{3}\right)}{4-3}=5\left(2-\sqrt{3}\right)\)
3 tháng 7 2020
\(\frac{5}{\sqrt{5}}=\frac{5\sqrt{5}}{\sqrt{5}.\sqrt{5}}\frac{5\sqrt{5}}{5}=\sqrt{5}\)
31 tháng 8 2021
a: \(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
b: \(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
c: \(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{5}}{30}\)
31 tháng 8 2021
a)\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
b)\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)
c)\(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{20}}{60}=\dfrac{\sqrt{5}}{30}\)
HN
0
31 tháng 8 2021
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
m: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
AD
1
T
15 tháng 9 2018
\(=\frac{\left(\sqrt{x}-\sqrt{4y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}+\frac{3x.\left(x-\sqrt{xy}\right)}{\left(x+\sqrt{xy}\right).\left(x-\sqrt{xy}\right)}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3x.\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x^2-xy}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3x\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x.\left(x-y\right)}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)}{x-y}+\frac{3\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\)
\(=\frac{\left(\sqrt{x}-2.\sqrt{y}\right).\left(\sqrt{x}-\sqrt{y}\right)+3.\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}-2.\sqrt{y}+3.\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=1\)
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) \(\frac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\frac{1}{3\sqrt{2}+2\sqrt{2}-2\sqrt{2}}=\frac{1}{3\sqrt{2}}=\frac{\sqrt{2}}{3\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{2}}{6}\)
Ta có: \(\sqrt{18}-\frac{1}{3}\sqrt{72}-\sqrt{8}+\frac{2-3\sqrt{2}}{3-\sqrt{2}}\)
\(=3\sqrt{2}-\frac{6\sqrt{2}}{3}-2\sqrt{2}+\frac{\left(3+\sqrt{2}\right)\left(2-3\sqrt{2}\right)}{9-2}\)
\(=3\sqrt{2}-2\sqrt{2}-2\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)