Question

Rút gọn:
\(A=\dfrac{2}{x-1}\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}\)

Answer 1

\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)

\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)

Answer 2

a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)

\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)

\(=\dfrac{1}{x}\)

b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)

\(=\dfrac{3x+6}{x-2}\)