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) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=0,2.10.\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\dfrac{\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}\sqrt{200}}{\dfrac{1}{8}}=\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}.10\sqrt{2}\right).8\)
\(=4\sqrt{\dfrac{1}{2}}-12\sqrt{2}+64\sqrt{2}=2\sqrt{4.\dfrac{1}{2}}+52\sqrt{2}=2\sqrt{2}+52\sqrt{2}=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)
\(=2\left|\sqrt{2}-3\right|+\left|-3\right|.\sqrt{2}-5\sqrt{1}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5\)
\(=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\left(4-6+2\sqrt{5}\right)-\sqrt{5}=-2+\sqrt{5}\)
b) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}.\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}.\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}.\sqrt{2}+\dfrac{4}{5}.10.\sqrt{2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}.\sqrt{2}+8.\sqrt{2}\right):\dfrac{1}{8}=\sqrt{2}\left(\dfrac{1}{4}-\dfrac{3}{2}+8\right).8\)
\(=\sqrt{2}.\dfrac{27}{4}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)
\(=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5.1^2=6-2\sqrt{2}+3\sqrt{2}-5\)
\(=1+\sqrt{2}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=\sqrt{5}-2\)
b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=\dfrac{1}{5}\cdot10\cdot\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
c) Ta có: \(\left(\dfrac{1}{2}\cdot\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\cdot\sqrt{2}+\dfrac{4}{5}\cdot\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3\sqrt{2}}{2}+\dfrac{4}{5}\cdot10\sqrt{2}\right)\cdot8\)
\(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+\dfrac{32\sqrt{2}}{4}\right)\cdot8\)
\(=2\cdot27\sqrt{2}\)
\(=54\sqrt{2}\)
d) Ta có: \(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\cdot\left(-3\right)^2}-5\cdot\sqrt{\left(-1\right)^4}\)
\(=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5\)
\(=6-2\sqrt{2}+3\sqrt{2}-5\)
\(=\sqrt{2}+1\)
