Trục căn thức ở mẫu : a . 6/5√8 ; b. 7/5+2√3 ; c. 6/√7 - √5 Giúp mik vs ạ
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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2021
\(\dfrac{5}{3\sqrt{8}}=\dfrac{5}{6\sqrt{2}}=\dfrac{5.\sqrt{2}}{6.2}=\dfrac{5\sqrt{2}}{12}\)
18 tháng 12 2020
1) \(5\sqrt{8}-\dfrac{7}{2}\sqrt{72}+6\sqrt{\dfrac{1}{2}}\\ =5.\sqrt{4^2.\dfrac{1}{2}}-\dfrac{7}{2}.\sqrt{12^2.\dfrac{1}{2}}+6.\sqrt{\dfrac{1}{2}}=\left(5.4+\dfrac{7}{2}.12+6\right)\sqrt{\dfrac{1}{2}}\\ =68\sqrt{\dfrac{1}{2}}\)
2) \(\dfrac{6}{\sqrt{5}-1}=\dfrac{6.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{6\left(\sqrt{5}+1\right)}{5-1}\\ =\dfrac{6\left(\sqrt{5}+1\right)}{4}=\dfrac{3.\left(\sqrt{5+1}\right)}{2}\)
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
\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)
\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)
\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
PT
1
XT
1
DT
1
14 tháng 9 2017
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{2}\)
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: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)
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) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)
\(=3\sqrt{7}+3\sqrt{5}\)