Question

help me

Answer 1

Ta có : \(\frac{\sqrt{4m^2-8mn+4n^2}}{3n-3m}=\frac{\sqrt{4\left(m^2-4mn+n^2\right)}}{3\left(n-m\right)}\)

\(=\frac{2\sqrt{\left(m-n\right)^2}}{3\left(n-m\right)}=\frac{2\left|m-n\right|}{3\left(n-m\right)}=\frac{2\left(n-m\right)}{3\left(n-m\right)}=\frac{2}{3}\)(vì m<n)

Answer 2

\(\frac{\sqrt{4m^2-8mn+4n^2}}{3n-3m}=\frac{\sqrt{4\left(m^2-4mn+m^2\right)}}{3\left(n-m\right)}=\frac{\sqrt{4\left(m-n\right)^2}}{3\left(n-m\right)}=\frac{2\left(m-n\right)}{3\left(n-m\right)}=-\frac{2}{3}\)

Answer 3

\(=\frac{\left|m-n\right|}{3\left(n-m\right)}\) \(=\mp\frac{1}{3}\)