Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\left[25-\left(2\sqrt{6}\right)^2\right]\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=1\)
Bạn tham khảo
https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764
Lời giải:
Biểu thức \(=\frac{(5+2\sqrt{6})(25+24-2\sqrt{25.24})\sqrt{3+2-2\sqrt{3.2}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(\sqrt{25}-\sqrt{24})^2.\sqrt{(\sqrt{3}-\sqrt{2})^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})^2(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)
\(=\frac{(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}=1\)
\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
- \(5-2\sqrt{6}=3-2\sqrt{2}\cdot\sqrt{3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\Rightarrow\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
- Tương tự \(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
- Tử số: \(TS=\left(\sqrt{3}+\sqrt{2}\right)^2\left(49-20\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)=\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\)
\(=49\sqrt{3}+49\sqrt{2}-20\cdot3\sqrt{2}-20\cdot2\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
- Vậy C = 1.
\(\left(\sqrt{2}+\sqrt{3}\right)^2\sqrt{49-20\sqrt{6}}=\left(\sqrt{2}+\sqrt{3}\right)^2.\sqrt{\left(2\sqrt{6}-5\right)^2}\)
\(=\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-24=1\)
\(\left(\sqrt{2}+\sqrt{3}\right)^2.\sqrt{49-20\sqrt{6}}=\left(\sqrt{2}+\sqrt{3}\right)\sqrt{25-20\sqrt{6}+24}\)
\(=\left(\sqrt{2}+\sqrt{3}\right).\sqrt{\left(5-2\sqrt{6}\right)^2}=\left(\sqrt{2}+\sqrt{3}\right).\left|5-2\sqrt{6}\right|\)
\(=\left(\sqrt{2}+\sqrt{3}\right)\left(5-2\sqrt{6}\right)=5\sqrt{2}+5\sqrt{3}-2\sqrt{12}-2\sqrt{18}\)
\(=5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}=\sqrt{3}-\sqrt{2}\)