Question

Rút gọn biểu thức A = \(\sqrt{\dfrac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)

Answer 1

đk: x khác 0

A = \(\sqrt{\dfrac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

= \(\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

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

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

TH1: x \(\ge2\)

A = \(\dfrac{x^2+3}{x}+x-2\)

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

TH2: \(0< x< 2\)

A = \(\dfrac{x^2+3}{x}-x+2\)

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

TH3: x < 0

A = \(\dfrac{x^2+3}{-x}-x+2\)

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