\(\frac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\)
K
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S
3 tháng 7 2019
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
S
3 tháng 7 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)
NV
16 tháng 7 2016
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
a)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2(\sqrt{6}+2+\sqrt{6}-2)}{(\sqrt{6}-2)(\sqrt{6}+2)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
b)
\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)
\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
c)
\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left[\frac{\sqrt{2}(\sqrt{3}-1)}{1-\sqrt{3}}-\sqrt{5}\right].(\sqrt{5}-\sqrt{2})\)
\(=(-\sqrt{2}-\sqrt{5})(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)
\(=-(5-2)=-3\)
d)
\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)
Đề bài là gì zậy bạn ?
Mk cx ko hiểu.
Mong bạn giúp đỡ, thanks nhiều ạ !
À ừm, sorry bạn .
Bài làm
\(\frac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}-\sqrt{5}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}-\sqrt{3}\right)^2-5}=\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{2+3-2\sqrt{6}-5}\)
\(=\frac{\sqrt{3}+\sqrt{5}-\sqrt{2}}{2\sqrt{6}}=\frac{\sqrt{18}+\sqrt{30}-\sqrt{12}}{12}\)