Question

Rút gọn

A= \(\sqrt{x^2}-\sqrt{x^2-4x+4};x\ge2\)

B=\(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9};x< 3\)

Answer 1

\(A=\sqrt{x^2}-\sqrt{x^2-4x+4}\)

\(\Leftrightarrow A=|x|-\sqrt{\left(x-2\right)^2}\)

\(\Leftrightarrow A=x-|x-2|=x-x+2=2\)

Answer 2

A = \(\sqrt{x^2}-\sqrt{x^2-4x+4}=\sqrt{x^2}-\sqrt{\left(x-2\right)^2}=\left|x\right|-\left|x-2\right|=x-x+2=2\)(vì  \(x\ge2\))

B = \(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=\left|x-3\right|-\left|x+3\right|=3-x+x+3=6\)(vì x < 3)

Answer 3

\(B=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)

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

\(\Leftrightarrow B=|x-3|-|x+3|=-x+3-x-3=0\)