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XT
1
LC
2
24 tháng 8 2023
a: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)
b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)
c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
a) \(\dfrac{6}{5\sqrt{8}}\)
\(=\dfrac{6}{5\cdot2\sqrt{2}}\)
\(=\dfrac{6}{10\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{10}\)
b) \(\dfrac{7}{5+2\sqrt{3}}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)
\(=\dfrac{35-14\sqrt{3}}{13}\)
c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)
\(=3\sqrt{7}+3\sqrt{5}\)
PL
0
7 tháng 10 2020
\(=\frac{6\sqrt{2}\left(3\sqrt{7}-5\sqrt{2}\right)}{2\left(3\sqrt{7}-5\sqrt{2}\right)}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\)
NT
0
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)
\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)
25 tháng 7 2023
\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)
\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)
\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)
a) `-2/(3\sqrt11) = (-2\sqrt11)/(3\sqrt11 .\sqrt11) =(-2\sqrt11)/(3.11)=(-2\sqrt11)/33`
b) `3/(\sqrt7+4) = (3.(\sqrt7-4))/((\sqrt7+4)(\sqrt7-4))`
`=(3.(\sqrt7-4))/((\sqrt7)^2-4^2)`
`=(3.(\sqrt7-4))/(-9)`
`=(4-\sqrt7)/3`
\(\dfrac{-2}{3\sqrt{11}}=\dfrac{-2\sqrt{11}}{33}\)
\(\dfrac{3}{4+\sqrt{7}}=\dfrac{12-3\sqrt{7}}{7}\)