a)\(\dfrac{3\sqrt{m}-1}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}-\dfrac{\sqrt{m}+1}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}\cdot\sqrt{m}\left(\sqrt{m}+1\right)=\dfrac{2\sqrt{m}-2}{\sqrt{m}-1}\cdot\sqrt{m}=\dfrac{2\left(\sqrt{m}-1\right)}{\sqrt{m}-1}\cdot\sqrt{m}=2\sqrt{m}\)
b)
Ta có :m=4-2\(\sqrt{3}\)
\(\Rightarrow\)\(\sqrt{m}\) =\(\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}-1\right)^2}\)
= \(\sqrt{3}\)-1
Thay vào ta có 2(\(\sqrt{3}-1\)) = 2\(\sqrt{3}\)-2