Bài $2$
$M=\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}$
$M=\sqrt{3-2\sqrt{3}+1}-\dfrac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}-\dfrac{\sqrt{3}(1-\sqrt{3}}{1-\sqrt{3}}$
$M=\sqrt{(\sqrt{3}-1)^2}-\dfrac{2(\sqrt{3}-1)}{3-1}-\sqrt{3}$
$M=\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}$
$M=-\sqrt{3}$
$N=\dfrac{3}{\sqrt{2}-1}-\dfrac{3\sqrt{6}-3\sqrt{10}}{\sqrt{3}-\sqrt{5}}$
$N=\dfrac{3(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}-\dfrac{3\sqrt{2}(\sqrt{3}-\sqrt{5})}{\sqrt{3}-\sqrt{5}}$
$N=\dfrac{3\sqrt{2}+3}{2-1}-3\sqrt{2}$
$N=3\sqrt{2}+3-3\sqrt{2}$
$N=3$
Bài $3$
$a)ĐKXĐ: x\ge1$
$\sqrt{x-1}=1$
`<=>`$x-1=1$
`<=>`$x=2(t/m)$
Vậy phương trình có nghiệm $x=2$
$b)ĐKXĐ: x\ge-\dfrac{1}{2}$
$2\sqrt{2x+1}=3$
`<=>`$\sqrt{2x+1}=\dfrac{3}{2}$
`<=>`$2x+1=\dfrac{9}{4}$
`<=>`$2x=\dfrac{5}{4}$
`<=>`$x=\dfrac{5}{8}(t/m)$
Vậy phương trình có nghiệm $x=\dfrac{5}{8}$
