a) \(\dfrac{6}{5}\sqrt{50}-3\sqrt{45}-2\sqrt{18}+5\sqrt{20}=\dfrac{6}{5}\sqrt{25.2}-3\sqrt{9.5}-2\sqrt{9.2}+5\sqrt{4.5}\)
\(=6\sqrt{2}-9\sqrt{5}-6\sqrt{2}+10\sqrt{5}=\sqrt{5}\)
b) \(\sqrt{\dfrac{\left(\sqrt{2}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{7}{2}+2\sqrt{3}}=\dfrac{\left|\sqrt{2}-\sqrt{3}\right|}{\sqrt{2}}-\sqrt{\dfrac{7+4\sqrt{3}}{2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\sqrt{\dfrac{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\sqrt{\dfrac{\left(2+\sqrt{3}\right)^2}{2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\dfrac{\left|2+\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\dfrac{2+\sqrt{3}}{\sqrt{2}}=\dfrac{-2-\sqrt{2}}{\sqrt{2}}=-\sqrt{2}-1\)
a) Ta có: \(\dfrac{6}{5}\sqrt{50}-3\sqrt{45}-2\sqrt{18}+5\sqrt{20}\)
\(=\dfrac{6}{5}\cdot5\sqrt{2}-3\cdot3\sqrt{5}-2\cdot3\sqrt{2}+5\cdot2\sqrt{5}\)
\(=6\sqrt{2}-9\sqrt{5}-6\sqrt{2}+10\sqrt{5}\)
=\(\sqrt{5}\)
b) Ta có: \(\sqrt{\dfrac{\left(\sqrt{2}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{7}{2}+2\sqrt{3}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\dfrac{2+\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-2-\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{2}-1\)
