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1) Cách 1 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(M=\sqrt{9-6\sqrt{2}+2}+\sqrt{9+6\sqrt{2}+2}\)
\(M=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(M=\left|3-\sqrt{2}\right|+\left|3+\sqrt{2}\right|\)
\(M=3-\sqrt{2}+3+\sqrt{2}=6\)
Cách 2 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(\Rightarrow M^2=11-6\sqrt{2}+2\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}+11+6\sqrt{2}\)
\(\Leftrightarrow M^2=22+2.7=36\)
\(\Leftrightarrow M=6\left(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}>0\right)\)
2)
\(A=53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{8-4\sqrt{2}+1}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\left|2\sqrt{2}-1\right|}\)
\(\Leftrightarrow A=53-20\sqrt{4+2\sqrt{2}-1}\)
\(\Leftrightarrow A=53-20\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow A=53-20\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow A=53-20\left(\sqrt{2}+1\right)\)
\(\Leftrightarrow A=53-20\sqrt{2}-20=33-20\sqrt{2}\)
3)
\(M=\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(M=\sqrt{3-\sqrt{5}}.\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\)
\(M=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(M=2\sqrt{2}.\sqrt{3-\sqrt{5}}\left(\sqrt{5}+1\right)\)
\(\Rightarrow M^2=8.\left(3-\sqrt{5}\right).\left(5+2\sqrt{5}+1\right)\)
\(\Leftrightarrow M^2=\left(24-8\sqrt{5}\right)\left(6+2\sqrt{5}\right)\)
\(\Leftrightarrow M^2=144+48\sqrt{5}-48\sqrt{5}-80\)
\(\Leftrightarrow M^2=64\Leftrightarrow M=8\left(\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)>0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right)\)
\(=0\)
b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{8}-2\sqrt{5}-6\sqrt{5}\right)\)
\(=\sqrt{\sqrt{3}}\left(4\sqrt{2}-8\sqrt{5}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\sqrt{\sqrt{3}}\cdot4\sqrt{5}-2\cdot\sqrt{\sqrt{3}}\cdot\sqrt{5}-3\cdot\sqrt{\sqrt{3}}\cdot2\sqrt{5}\)
\(=2\sqrt{\sqrt{3}}\left(4\sqrt{5}-\sqrt{5}-3\sqrt{5}\right)=0\)
b: \(=2\cdot2\sqrt{2}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(2\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)=2\sqrt{\sqrt{3}}\cdot\left(2\sqrt{2}-4\sqrt{5}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath