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5 tháng 8 2021
\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
AO
2
5 tháng 6 2017
\(\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}=\left(5+2\sqrt{6}\right)^3\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{2}\right)^6\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)^5.1=\left(\sqrt{3}+\sqrt{2}\right)^5\)
5 tháng 6 2017
$\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}
=\left(5+2\sqrt{6}\right)^3\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}
=\left(\sqrt{3}+\sqrt{2}\right)^6\left(\sqrt{3}-\sqrt{2}\right)
=\left(\sqrt{3}+\sqrt{2}\right)^5.1
=\left(\sqrt{3}+\sqrt{2}\right)^5$
NT
1
12 tháng 7 2018
\(=\left(2+2\sqrt{2}\cdot\sqrt{3}+3\right)\left(25-2\cdot5\cdot\sqrt{24}+24\right)\sqrt{3-2\sqrt{3}\sqrt{2}+2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)^2\left(5-\sqrt{24}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\)
\(=9\sqrt{3}-11\sqrt{2}\)
AH
Akai Haruma
Giáo viên
3 tháng 7 2021
Bạn tham khảo
https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764
TA
30 tháng 6 2017
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left[25-\left(2\sqrt{6}\right)^2\right]\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=1\)
AH
Akai Haruma
Giáo viên
7 tháng 8 2019
Lời giải:
Biểu thức \(=\frac{(5+2\sqrt{6})(25+24-2\sqrt{25.24})\sqrt{3+2-2\sqrt{3.2}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(\sqrt{25}-\sqrt{24})^2.\sqrt{(\sqrt{3}-\sqrt{2})^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})^2(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}=1\)
Y
5 tháng 7 2019
pt nào z bn?
\(=\left(3+2\sqrt{3\cdot2}+2\right)\left(25-2\cdot5\cdot2\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^4\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left[\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\right]^2\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(=\left(\sqrt{3}-\sqrt{2}\right)^3\)
BL
10 tháng 7 2017
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
BL
10 tháng 7 2017
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
\(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2.\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^4.\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right) ^3\)
\(=\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(=\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(49-20\sqrt{6}\right)\)
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