Question

Rút gọn: \(\frac{\sqrt{X^2-4x+4}}{X-2}\) với X<2

\(\frac{\sqrt{9X^2-6X+1}}{9X^2-1}\) với X>1/3

Answer 1

a/ \(\frac{\sqrt{x^2-4x+4}}{x-2}=\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-2\right|}{x-2}\)

có x<2\(\Rightarrow\left|x-2\right|=2-x\)

\(\Rightarrow\frac{2-x}{x-2}\)

b/ \(\frac{\sqrt{9x^2-6x+1}}{9x^2-1}=\frac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}\)

Có x>\(\frac{1}{3}\Rightarrow\left|3x-1\right|=3x-1\)

\(\Rightarrow\frac{3x-1}{\left(3x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)