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\(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2.\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^4.\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right) ^3\)
\(=\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(49-20\sqrt{6}\right)\)
\(\)
\(=\left(2+2\sqrt{2}\cdot\sqrt{3}+3\right)\left(25-2\cdot5\cdot\sqrt{24}+24\right)\sqrt{3-2\sqrt{3}\sqrt{2}+2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)^2\left(5-\sqrt{24}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\)
\(=9\sqrt{3}-11\sqrt{2}\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
pt nào z bn?
\(=\left(3+2\sqrt{3\cdot2}+2\right)\left(25-2\cdot5\cdot2\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^4\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left[\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\right]^2\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(=\left(\sqrt{3}-\sqrt{2}\right)^3\)
Bạn tham khảo
https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764
\(\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}=\left(5+2\sqrt{6}\right)^3\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{2}\right)^6\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)^5.1=\left(\sqrt{3}+\sqrt{2}\right)^5\)
$\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}
=\left(5+2\sqrt{6}\right)^3\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}
=\left(\sqrt{3}+\sqrt{2}\right)^6\left(\sqrt{3}-\sqrt{2}\right)
=\left(\sqrt{3}+\sqrt{2}\right)^5.1
=\left(\sqrt{3}+\sqrt{2}\right)^5$