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\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)
\(=-15\sqrt{2}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
=\(\sqrt{4}.\sqrt{3}+2\sqrt{9}.\sqrt{3}+3\sqrt{25}.\sqrt{3}-9\sqrt{16}.\sqrt{3}\)
=\(2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
=\(\left(2+6+15-36\right)\sqrt{3}\)
=\(-13\sqrt{3}\)
@Nguyễn Thị Thu Sương :
\(\frac{\sqrt{3+\sqrt{15}}}{\sqrt{2}}=\sqrt{\frac{3+\sqrt{15}}{2}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{5-3}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{\sqrt{3}}{\sqrt{5}-\sqrt{3}}}\)
a) \(\left(\sqrt{12}-\sqrt{27}+\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-3\sqrt{3}+\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{3}\left(2-3+1\right):\sqrt{3}\)
\(=0:\sqrt{3}=0\)
b) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
\(=\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
\(=\frac{5\sqrt{3}}{\sqrt{3}\cdot\sqrt{5}}+\frac{3\sqrt{5}}{\sqrt{3}\cdot\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{3}\)
Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)
Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)
Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\in N\)
Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)
Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)
Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)
\(\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3\sqrt{5}-3\sqrt{2}}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3.\left(\sqrt{5}-\sqrt{2}\right)}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3}.\sqrt{\sqrt{5}-\sqrt{2}}}\)
\(=\frac{(\sqrt{\sqrt{5}+\sqrt{2}})^2}{\sqrt{3}.\sqrt{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{5-2}}\)
\(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{3}}=\frac{\sqrt{5}+\sqrt{2}}{3}\)