Giải bài 2 trang 131 – SGK Toán lớp 9 tập 2

Ta có:

$\begin{aligned} M&=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}} \\ & =\sqrt{2-2\sqrt{2}+1}-\sqrt{4+4\sqrt{2}+2} \\ & =\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}-\sqrt{{{\left( 2+\sqrt{2} \right)}^{2}}} \\ & =\sqrt{2}-1-\left( 2+\sqrt{2} \right) \\ & =-3 \\ \end{aligned}$

$\begin{aligned} N\sqrt{2} &=\sqrt{2}.\sqrt{2+\sqrt{3}}+\sqrt{2}.\sqrt{2-\sqrt{3}} \\ & =\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}} \\ & =\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1} \\ & =\sqrt{{{\left( \sqrt{3}+1 \right)}^{2}}}+\sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}} \\ & =\sqrt{3}+1+\left( \sqrt{3}-1 \right) \\ & =2\sqrt{3} \\ \end{aligned} \\ \Rightarrow N=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}$