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25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
4 tháng 9 2017
\(=\frac{2\sqrt{5\cdot3}-2\sqrt{5\cdot2}+\sqrt{2\cdot3}-\sqrt{3\cdot3}}{2\sqrt{5}-2\sqrt{2\cdot5}-\sqrt{3}+\sqrt{2\cdot3}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}{-\left(\sqrt{2}-1\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
22 tháng 7 2023
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
HN
1
6 tháng 2 2021
Ta có: \(A=\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)
\(=\dfrac{-2\sqrt{6}\left(5-\sqrt{6}\right)}{5-\sqrt{6}}-\sqrt{\dfrac{2}{3}\cdot9}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)
\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)
\(=-3\sqrt{6}+3\sqrt{6}+3\)
=3
PT
2
17 tháng 7 2016
\(x=\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\Rightarrow x^3=10+3.\sqrt[3]{5^2-\left(2\sqrt{6}\right)^2}.x\Leftrightarrow x^3=10+3x\)
\(\Rightarrow\frac{x^3-10}{x}=3\Leftrightarrow x^2-\frac{10}{x}=3\Rightarrow B=3\)
17 tháng 7 2016
Đặt \(a=\sqrt[3]{5+2\sqrt{6}};b=\sqrt[3]{5-2\sqrt{6}}\).Ta có :
B = x2 - \(\frac{10}{x}=\frac{x^3-10}{x}=\frac{\left(a+b\right)^3-10}{x}=\frac{a^3+b^3+3ab\left(a+b\right)-10}{x}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}+3x.\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}-10}{x}\)\(=\frac{3x.\sqrt[3]{5^2-\left(2\sqrt{6}\right)^2}}{x}=3.\sqrt[3]{25-24}=3.\sqrt[3]{1}=3\)
VT
30 tháng 9 2019
a)=\(\sqrt{3-\sqrt{5}}\).\(\sqrt{3+\sqrt{5}}\).\(\sqrt{2}\)(\(\sqrt{5}\)-\(1\))\(\sqrt{3+\sqrt{5}}\)=2\(\sqrt{2}\) \(\sqrt{\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\) .\(\sqrt{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\)\(\sqrt{8}\) =8
b)A2=8+2 căn[\(\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)\)]=8+2\(\sqrt{6-2\sqrt{5}}\)=8+2(\(\sqrt{5}\)-1)=6+2\(\sqrt{5}\)=(\(\sqrt{5}+1\))2 =>A=\(\sqrt{5}\)+1
c)C=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right).\sqrt{2}.\left(\sqrt{5}-1\right)^2.\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2.\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2.\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)
\(=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)