c: \(\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{3}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{3}{\sqrt{3}-1}\)
\(=\dfrac{2\sqrt{3}-2-3+\sqrt{3}+3}{\sqrt{3}-1}\)
\(=\dfrac{3\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{7+\sqrt{3}}{2}\)
1) \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{2.\left(-5\right)^2}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}\)
2) \(\dfrac{\sqrt{270}}{\sqrt{30}}-\sqrt{1,8}.\sqrt{20}=\dfrac{3\sqrt{30}}{\sqrt{30}}-\sqrt{6}=3-\sqrt{6}\)
Bài 2:
a) \(=6\sqrt{2}-5\sqrt{2}+6\sqrt{3}=\sqrt{2}+6\sqrt{3}\)
b) \(=3\sqrt{2}-\sqrt{2}-5\sqrt{2}=-3\sqrt{2}\)
c) \(=\dfrac{2\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{3\left(\sqrt{3}+1\right)}{3-1}-\sqrt{3}\)
\(=2+\dfrac{3\sqrt{3}+3}{2}-\sqrt{3}=\dfrac{4+3\sqrt{3}+3-2\sqrt{3}}{2}=\dfrac{7-\sqrt{3}}{2}\)
\(1,=3-\sqrt{2}+5\sqrt{2}=3-4\sqrt{2}\\ 2,=\sqrt{\dfrac{270}{30}}-\sqrt{1,8\cdot20}=\sqrt{9}-\sqrt{36}=3-6=-3\\ a,=6\sqrt{2}-5\sqrt{2}+6\sqrt{3}=6\sqrt{3}+\sqrt{2}\\ b,=3\sqrt{2}-\sqrt{2}-5\sqrt{2}=-3\sqrt{2}\\ c,=\dfrac{2\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{3\left(\sqrt{3}+1\right)}{2}-\sqrt{3}\\ =2+\dfrac{3\sqrt{3}+3}{2}-\sqrt{3}=\dfrac{4+3\sqrt{3}+3-2\sqrt{3}}{2}=\dfrac{7+\sqrt{3}}{2}\)
