Thu gọn biểu thức :
\(P=\frac{1}{\sqrt{2}-\sqrt{3}}.\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
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20 tháng 8 2020
Ta có: \(E=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
\(=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)
\(=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\)
\(=-\frac{1}{\sqrt{3}-\sqrt{2}}\cdot\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\)
\(=-\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\cdot\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)
\(=-\sqrt{\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}}\)
\(=-\sqrt{\frac{1}{3-2}}=-1\)
ML
30 tháng 5 2016
Q=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3-2}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\left(\sqrt{3}-\sqrt{2}\right)^{ }\)=-1
6 tháng 7 2019
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
6 tháng 7 2019
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
5 tháng 8 2019
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
LT
8 tháng 8 2019
1) \(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}\)
= \(\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\)
= 6+3
=9
2) \(\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\frac{3+\sqrt{3}}{\sqrt{3}}\)
= \(\frac{-\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}\)
= \(-\sqrt{3}+\sqrt{3}+1\)
=1
3)\(\frac{2-\sqrt{2}}{1-\sqrt{2}}+\frac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
= \(\frac{-\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)
= \(-2\sqrt{2}\)
12 tháng 10 2018
\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{4}:\left(\frac{2\sqrt{2+\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{2\sqrt{2+\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{4}:\left(\frac{\sqrt{4+2\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)
\(=\frac{\sqrt{3}-1}{4}.\frac{4\sqrt{6}}{\sqrt{2}\left(\sqrt{9}+2\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}+1\right)^2}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}\)............