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a, \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(2\left(-5\right)\right)^2}\)
\(=\left|3-\sqrt{2}\right|+\sqrt{\left(-10\right)^2}\)
\(=3-\sqrt{2}+\left|-10\right|\)
\(=3-\sqrt{2}+10\)
\(=13-\sqrt{2}\)
b, \(\dfrac{\sqrt{270}}{\sqrt{30}}-\sqrt{1,8}.\sqrt{20}\)
\(=\sqrt{9}-\sqrt{1,8.20}\)
\(=3-\sqrt{36}\)
\(=3-6\)
\(=-3\)
a,
\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
\(=6\sqrt{6}^2-12\sqrt{9.2}+15\sqrt{4.3}-\dfrac{3}{4}\sqrt{16.3}\)
= \(6.6-12.3\sqrt{2}+15.2\sqrt{3}-\dfrac{3}{4}.4\sqrt{3}\)
\(=36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\)
\(=36-36\sqrt{2}+27\sqrt{3}\)
\(\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(\sqrt{18}-\sqrt{20}+2\sqrt{2}\right)\)
\(=\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(3\sqrt{2}-2\sqrt{5}+2\sqrt{2}\right)\)
\(=\left(5\sqrt{2}-\sqrt{5}\right)\left(5\sqrt{2}-2\sqrt{5}\right)\)
\(=50-10\sqrt{10}-5\sqrt{10}+10\)
\(=60-15\sqrt{10}\)
\(\left(1+\sqrt{2}-\sqrt{5}\right)\left(1+\sqrt{2}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{2}\right)^2-5\)
\(=1+2\sqrt{2}+2-5\)
\(2\sqrt{2}-2\)
Câu a : \(\left(\sqrt{80}+\sqrt{20}\right):\sqrt{45}=\sqrt{80}:\sqrt{45}+\sqrt{20}:\sqrt{45}=\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{4}{9}}=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
Câu b : \(\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{18}+\sqrt{27}\right)=\sqrt{54}+\sqrt{81}-\sqrt{36}-\sqrt{54}=\sqrt{81}-\sqrt{36}=9-6=3\)
Câu c : \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{15+3}}=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}-\dfrac{6}{\sqrt{18}}\)
\(=\sqrt{15}-\dfrac{6}{\sqrt{18}}=\dfrac{\sqrt{270}-6}{3\sqrt{2}}=\dfrac{3\sqrt{30}-6}{3\sqrt{2}}=\dfrac{3\left(\sqrt{30}-6\right)}{3\sqrt{2}}=\dfrac{\sqrt{30}-2}{\sqrt{2}}=\sqrt{15}-\sqrt{2}\)
\(a.\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\dfrac{\left(2+\sqrt{3}\right)\sqrt{3-2\sqrt{3}+1}}{\sqrt{3+2\sqrt{3}+1}}=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{2\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3-1}=4-3=1\)
\(b.\dfrac{\left(\sqrt{5}-1\right)^3}{\sqrt{5}-2}=\dfrac{5\sqrt{5}-15+3\sqrt{5}-1}{\sqrt{5}-2}=\dfrac{8\sqrt{5}-16}{\sqrt{5}-2}=\dfrac{8\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=8\)
\(c.\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right]=2\left(3+1+3\right)=2.7=14\)
\(d.\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=\dfrac{2}{2}=1\)
câu a
câu b