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\(\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\left(3+2\sqrt{2}\right)\sqrt{3-2\sqrt{2}}\)
\(=\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\sqrt{\left(3+2\sqrt{2}\right)}\sqrt{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\)
\(=\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\sqrt{\left(3+2\sqrt{2}\right)}\)
\(=\left(\left(2-\sqrt{2}\right)^2-3\right)\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left(3-4\sqrt{2}\right)\left(\sqrt{2}+1\right)=-\sqrt{2}-5\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)
b)\(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
c)\(\sqrt{\left(4-\sqrt{17}\right)^2}=\left|4-\sqrt{17}\right|=\sqrt{17}-4\)
d)\(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+\left|2-\sqrt{3}\right|=2\sqrt{3}+2-\sqrt{3}\)
Ta xét:
\(a=\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\)
\(a=\left(2-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\)
\(a=6-4\sqrt{2}-3\)
\(a=3-4\sqrt{2}\)
và
\(b=\left(3+2\sqrt{2}\right)\sqrt{3-2\sqrt{2}}\)
\(b=\sqrt{3+2\sqrt{2}}\cdot\left(\sqrt{3+2\sqrt{2}}\cdot\sqrt{3-2\sqrt{2}}\right)\)
\(b=\sqrt{3+2\sqrt{2}}\cdot\sqrt{3^2-\left(2\sqrt{2}\right)^2}\)
\(b=\sqrt{3+2\sqrt{2}}\cdot\sqrt{9-8}\)
\(b=\sqrt{3+2\sqrt{2}}\)
=> \(ab=\left(3-4\sqrt{2}\right)\sqrt{3+2\sqrt{2}}\)