Tính C = \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
(Đáp án bằng 1 nha các bạn).Mình Ko biết cách giải
\(\frac{\left(5+\sqrt{24}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}\)
Tính ghi rõ cách giải
\(\frac{\left(5+\sqrt{24}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2.\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}\)
\(=\frac{\left(25-24\right)\left(\sqrt{3}-\sqrt{2}\right)^2.\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{30}-11\sqrt{2}}\)\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)^3}{9\sqrt{30}-11\sqrt{2}}\)
Đến đây k biết làm
Tính \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left[25-\left(2\sqrt{6}\right)^2\right]\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=1\)
\(\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right).\left(25-20\sqrt{6}+24\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^2.\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(25-\left(2\sqrt{6}\right)^2\right).\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{1.\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=1\)
Học tốt
Tính
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)nhan\(\frac{\sqrt{5-2\sqrt{6}}}{1}\)
Tính:
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Bạn tham khảo
https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Rút gọn:
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Lời giải:
Biểu thức \(=\frac{(5+2\sqrt{6})(25+24-2\sqrt{25.24})\sqrt{3+2-2\sqrt{3.2}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(\sqrt{25}-\sqrt{24})^2.\sqrt{(\sqrt{3}-\sqrt{2})^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})^2(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}=1\)
Giúp mình với, ko cần làm hết đâu. Tính!
a)\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b)\(\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
c)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
d)\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2