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Kim Ngân Nguyễn Thị
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Chiharu
2 tháng 10 2019 lúc 21:47

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Lưu Như Ý
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Dương Thân Thùy
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Akai Haruma
8 tháng 7 2019 lúc 13:53

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)

\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)

\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)

Akai Haruma
8 tháng 7 2019 lúc 14:00

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)

\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)

Hiền Vũ Thu
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Nguyễn Quang Bảo
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Nguyễn Quang Bảo
14 tháng 7 2018 lúc 21:40

Giusp minh voi a

minh
14 tháng 7 2018 lúc 21:47

bạn vào wolfram alpha mà tính

Hoang Quoc Khanh
14 tháng 7 2018 lúc 21:54

Yêu cầu là gì vậy bạn?

shoppe pi pi pi pi
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shoppe pi pi pi pi
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shoppe pi pi pi pi
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svtkvtm
3 tháng 7 2019 lúc 15:31

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)

svtkvtm
3 tháng 7 2019 lúc 15:42

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)

Linh Nguyen
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Nguyễn Lê Phước Thịnh
27 tháng 8 2020 lúc 13:54

a) Ta có: \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)

\(=\frac{\sqrt{5}-2}{\sqrt{5}\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}+\frac{\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)

\(=\frac{\sqrt{5}-2-\sqrt{5}+\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)

\(=\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}\)

\(=\frac{1}{\sqrt{5}+2}\)

b) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

\(=\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{2}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}-\frac{2\sqrt{2}\cdot\left(\sqrt{3}+2\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)

\(=\frac{3\sqrt{2}+\sqrt{6}+\sqrt{2}\cdot\left(5+3\sqrt{3}\right)-2\sqrt{6}-4\sqrt{2}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)

\(=\frac{-\sqrt{2}-\sqrt{6}+5\sqrt{2}+3\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)

\(=\frac{4\sqrt{2}+2\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)

\(=\frac{2\sqrt{2}\cdot\left(2+\sqrt{3}\right)}{\sqrt{3}\cdot\sqrt{2}\cdot\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)

\(=\frac{2}{3+\sqrt{3}}\)

Vy Nguyễn Đặng Khánh
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Nguyen
4 tháng 7 2019 lúc 20:04

\(A=\frac{1}{\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)}+\frac{1}{\sqrt{3}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)}+\frac{1}{\sqrt{5}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)}\)

\(=\frac{1}{\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)\)

=1