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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) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
b)\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
\(\Leftrightarrow\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
\(\Leftrightarrow\sqrt{m-1}+1+\sqrt{m-1}-1\Leftrightarrow2\sqrt{m-1}\)
Câu 1 phá từng lớp ra :VD\(9+4\sqrt{2}\) =\((\sqrt{2}+2)^2\)
Câu 2:m+2\(\sqrt{m-1}\) =m-1+1+2\(\sqrt{m-1}\) =\((\sqrt{m-1} -1)^2 \)
\(A=\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
\(A=\dfrac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+2\sqrt{2}\right)}+\dfrac{\sqrt{2}.\left(\sqrt{3}+3\right)}{\sqrt{2}.\left(\sqrt{2+\sqrt{3}}-2\sqrt{2}\right)}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{4-2\sqrt{3}}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{4+2\sqrt{3}}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{3}-1+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}+1-4}\)
\(A=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{6}\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}+\dfrac{\sqrt{3}\left(\sqrt{2}+\sqrt{6}\right)}{\sqrt{3}\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}}{1+\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{6}}{1-\sqrt{3}}=\dfrac{\left(\sqrt{2}-\sqrt{6}\right)\left(1-\sqrt{3}\right)+\left(\sqrt{2}+\sqrt{6}\right)\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}-\sqrt{6}+3\sqrt{2}+\sqrt{2}+\sqrt{6}+\sqrt{6}+3\sqrt{2}}{1-3}=\dfrac{8\sqrt{2}}{-2}=-4\sqrt{2}\)
* \(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\) \(=\dfrac{\sqrt{6+2.\sqrt{6}.\sqrt{5}+5}-\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}{\sqrt{5}}\)\(=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}}{\sqrt{5}}=\dfrac{2\sqrt{5}}{\sqrt{5}}=2\)
* \(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)
Đặt:\(y=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\Rightarrow y^2=4+\sqrt{15}+4-\sqrt{15}+2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=8+2=10\Rightarrow y=\sqrt{10}\)
Suy ra: \(C=\sqrt{12+4\sqrt{5}}-y=\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}-\sqrt{10}=\sqrt{10}+\sqrt{2}-\sqrt{10}=\sqrt{2}\)* \(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)+\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{1}=4\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{4-3}=1\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}.\)
\(=\sqrt{4+\sqrt{5.\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}=\dfrac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}=\dfrac{4+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}}{2\sqrt{2}}\) \(=\dfrac{4+2\sqrt{7}-2\sqrt{7}}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
1/
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{4-3}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
2/
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(5-2\right)=-3\)
#F.C
\(\sqrt{2+\sqrt{3}}\)\(\times\sqrt{2+\sqrt{2+\sqrt{3}}}\)\(\times\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\)\(\times\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
= \(\sqrt{2+\sqrt{3}}\)\(\times\sqrt{2+\sqrt{2+\sqrt{3}}}\)\(\times\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
= \(\sqrt{2+\sqrt{3}}\)\(\times\sqrt{2+\sqrt{2+\sqrt{3}}}\)\(\times\sqrt{2-\sqrt{2+\sqrt{3}}}\)
= \(\sqrt{2+\sqrt{3}}\)\(\times\sqrt{4-2-\sqrt{3}}\)
= \(\sqrt{2+\sqrt{3}}\)\(\times\sqrt{2-\sqrt{3}}\)
= \(\sqrt{4-3}\)
= 1