Bài 1:
a) Ta có: \(A=3\sqrt{8}-\sqrt{50}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=3\cdot2\sqrt{2}-5\sqrt{2}-\left(\sqrt{2}-1\right)\)
\(=6\sqrt{2}-5\sqrt{2}-\sqrt{2}+1\)
\(=1\)
Ta có: \(B=\dfrac{3}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{9x^2}}\)
\(=\dfrac{3}{x-1}\cdot\dfrac{\left|x-1\right|}{3x}\)
\(=\dfrac{-\left(x-1\right)}{x-1}\cdot\dfrac{1}{x}\)(Vì 0<x<1)
\(=-1\cdot\dfrac{1}{x}=-\dfrac{1}{x}\)
b) Để \(B=\dfrac{-2}{\sqrt{x}}\) thì \(\dfrac{-1}{x}=\dfrac{-2}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{2\sqrt{x}}{x}\)
Suy ra: \(2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
hay \(x=\dfrac{1}{4}\)(nhận)
Vậy: Để \(B=\dfrac{-2}{\sqrt{x}}\) thì \(x=\dfrac{1}{4}\)
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