\(1,\\ a,=\dfrac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)}}=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}}=\sqrt{\sqrt{a}-\sqrt{b}}\\ b,=\dfrac{\sqrt{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}}{\sqrt{\sqrt{x}+\sqrt{3}}}\cdot\dfrac{\sqrt{3}}{\sqrt{\sqrt{x}-\sqrt{3}}}\\ =\sqrt{3}\\ c,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\\ d,=5xy\cdot\dfrac{\left|5x\right|}{y^2}=\dfrac{-25x^2y}{y^2}=\dfrac{-25x^2}{y}\)
Bài 2:
a: Ta có: \(A=\left(3\sqrt{18}+2\sqrt{50}-4\sqrt{72}\right):8\sqrt{2}\)
\(=\left(9\sqrt{2}+10\sqrt{2}-24\sqrt{2}\right):8\sqrt{2}\)
\(=\dfrac{-5\sqrt{2}}{8\sqrt{2}}=-\dfrac{5}{8}\)
b: Ta có: \(B=\left(-4\sqrt{20}+5\sqrt{500}-3\sqrt{45}\right):\sqrt{5}\)
\(=\left(-8\sqrt{5}+50\sqrt{5}-9\sqrt{5}\right):\sqrt{5}\)
\(=49\)