Dạng 3: Biểu thức dưới dấu căn đưa được về hằng đẳng thức

$$\begin{gathered} A=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3} \\ =\sqrt{{\left(\sqrt{3}-1\right)}^2}-\sqrt{{\left(2+\sqrt{3}\right)}^2} \\ =\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right| \\ =\sqrt{3}-1-\left(2+\sqrt{3}\right)=-3 \end{gathered}$$

$A=\sqrt{6-2\sqrt{5}}=\sqrt{{\left(\sqrt{5}-1\right)}^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1$

$B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}=\sqrt{{\left(\sqrt{3}-1\right)}^2}=\left|\sqrt{3}-1\right|$

$C=\sqrt{19-8\sqrt{3}}=\sqrt{{\left(4-\sqrt{3}\right)}^2}=\left|4-\sqrt{3}\right|=4-\sqrt{3}$

$D=\sqrt{5-2\sqrt{6}}=\sqrt{{\left(\sqrt{3}-\sqrt{2}\right)}^2}=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}$

$A=\sqrt{4+2\sqrt{3}}=\sqrt{{\left(\sqrt{3}+1\right)}^2}=\sqrt{3}+1$

$B=\sqrt{8-2\sqrt{15}}=\sqrt{{\left(\sqrt{15}-1\right)}^2}=\sqrt{15}-1$

$C=\sqrt{9-4\sqrt{5}}=\sqrt{{\left(2-\sqrt{5}\right)}^2}=\sqrt{5}-2$

$D=\sqrt{7+\sqrt{13}}-\sqrt{7-\sqrt{13}}=\frac{1}{\sqrt{2}}\left(\sqrt{14+2\sqrt{13}}-\sqrt{14-2\sqrt{13}}\right)$

$=\frac{1}{\sqrt{2}}\left[\sqrt{{\left(\sqrt{13}+1\right)}^2}-\sqrt{{\left(\sqrt{13}-1\right)}^2}\right]=\sqrt{2}$

$E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}$

$=\sqrt{{(\sqrt{5}+1)}^2}-\sqrt{{(\sqrt{5}-1)}^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-\sqrt{5}+1=2$

$F=\sqrt{7-2\sqrt{10}}+\sqrt{20}+\frac{1}{2}\sqrt{8}=\sqrt{{\left(\sqrt{5}-\sqrt{2}\right)}^2}+2\sqrt{5}+\frac{1}{2}.2\sqrt{2}$

$=\left|\sqrt{5}-\sqrt{2}\right|+2\sqrt{5}+\sqrt{2}=\sqrt{5}-\sqrt{2}+2\sqrt{5}+\sqrt{2}=3\sqrt{5}$