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A\(\left(3-\sqrt{3}\right)\left(-2\sqrt{3}\right)+\left(3\sqrt{3}+1\right)^2\)=\(6-6\sqrt{3}+9+6\sqrt{3}+1\)
=16
B,\(\left(3\sqrt{5}-2\sqrt{3}\right)\sqrt{5}+\sqrt{60}\) =\(15-2\sqrt{15}+2\sqrt{15}=15\)
\(\sqrt{3}< 2;\sqrt{3}>1\)
\(\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\)
\(2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
\(b,\sqrt{5}< \sqrt{9}=3\)
\(\left|\sqrt{5}-3\right|-2\sqrt{5}+2\)
\(3-\sqrt{5}-2\sqrt{5}+2\)
\(5-3\sqrt{5}\)
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
b) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=2\sqrt{2}-3\)
a)\(\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\) (vì 2>\(√3\))
b) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\) (vì 3>\(2\sqrt{2}\))