1.1
\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|+|\sqrt{5}-1|\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
1.2
\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{2.3}+2}+\sqrt{3-2\sqrt{3.2}+2}\)
\(=\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2}=|\sqrt{3}+\sqrt{2}|+|\sqrt{3}-\sqrt{2}|\)
\(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
1.3
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{27-2\sqrt{27.8}+8}\)
\(=\sqrt{(3-\sqrt{6})^2}+\sqrt{(\sqrt{27}-\sqrt{8})^2}=|3-\sqrt{6}|+|\sqrt{27}-\sqrt{8}|\)
\(=3-\sqrt{6}+\sqrt{27}-\sqrt{8}=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
1.4
\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\sqrt{45-2\sqrt{45}+1}-\sqrt{20-2\sqrt{20.9}+9}\)
\(=\sqrt{(\sqrt{45}-1)^2}-\sqrt{(\sqrt{20}-\sqrt{9})^2}=|\sqrt{45}-1|-|\sqrt{20}-\sqrt{9}|\)
\(=\sqrt{45}-1-(\sqrt{20}-\sqrt{9})=3\sqrt{5}-1-2\sqrt{5}+3=\sqrt{5}+2\)
1.5: Trùng 1.4
1.6
\((3-\sqrt{2})\sqrt{11+6\sqrt{2}}=(3-\sqrt{2})\sqrt{9+2.3\sqrt{2}+2}\)
\(=(3-\sqrt{2})\sqrt{(3+\sqrt{2})^2}=(3-\sqrt{2})(3+\sqrt{2})=9-2=7\)
1.7
\((\sqrt{2}+1)\sqrt{3-2\sqrt{2}}=(\sqrt{2}+1)\sqrt{2-2\sqrt{2}+1}\)
\(=(\sqrt{2}+1)\sqrt{(\sqrt{2}-1)^2}=(\sqrt{2}+1)|\sqrt{2}-1|=(\sqrt{2}+1)(\sqrt{2}-1)=2-1=1\)
1.8
\((2+\sqrt{7})\sqrt{11-4\sqrt{7}}\)
\(=(2+\sqrt{7})\sqrt{4-2.2\sqrt{7}+7}\)
\(=(2+\sqrt{7})\sqrt{(2-\sqrt{7})^2}=(2+\sqrt{7})|2-\sqrt{7}|=(2+\sqrt{7})(\sqrt{7}-2)\)
\(=7-2^2=3\)
1.9
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{20-2.\sqrt{20.9}+9}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-\sqrt{9})^2}}=\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}\)
\(=\sqrt{5}-\sqrt{6-\sqrt{20}}=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{3-2\sqrt{3.2}+2}\)
\(=\sqrt{5}-\sqrt{(\sqrt{3}-\sqrt{2})^2}=\sqrt{5}-(\sqrt{3}-\sqrt{2})=\sqrt{5}-\sqrt{3}+\sqrt{2}\)
Bài 2:
\(A=(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}):\sqrt{3}=(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}):\sqrt{3}\)
\(=-5\sqrt{3}:\sqrt{3}=-5\)
\(B=\sqrt{3}-\sqrt{12}+\sqrt{27}=\sqrt{3}-2\sqrt{3}+3\sqrt{3}=(1-2+3)\sqrt{3}=2\sqrt{3}\)
\(C=\sqrt{27}-2\sqrt{12}-\sqrt{75}=3\sqrt{3}-4\sqrt{3}-5\sqrt{3}=-6\sqrt{3}\)
\(D=2\sqrt{3}+3\sqrt{27}-\sqrt{300}=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\)
Bài 1:
1) Ta có: \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
2) Ta có: \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)
\(=2\sqrt{3}\)
7) Ta có: \(\left(\sqrt{2}+1\right)\cdot\sqrt{3-2\sqrt{2}}\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\)
=2-1=1
Bài 2:
a) Ta có: \(A=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=2-5\cdot3+4\cdot2\)
\(=2-15+8=10-15=-5\)
b) Ta có: \(B=\sqrt{3}-\sqrt{12}+\sqrt{27}\)
\(=\sqrt{3}-2\sqrt{3}+3\sqrt{3}\)
\(=2\sqrt{3}\)
c) Ta có: \(C=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)
\(=3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\)
\(=-6\sqrt{3}\)
