Dạng 4: Rút gọn tổng hợp (sử dụng trục căn thức, hằng đẳng thức, phân tích thành nhân tử; …)

$A=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}+\frac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{5}+1}{\sqrt{5}+1}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=2$

$B=\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{6}+\sqrt{5}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+\left(\sqrt{6}-\sqrt{5}\right)$

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

$C=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\mathrm{...}+\frac{1}{\sqrt{99}+\sqrt{100}}$

$=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\mathrm{...}+\left(\sqrt{100}-\sqrt{99}\right)=9$

$$\begin{gathered} D=\frac{1}{2-\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\frac{1}{2-\sqrt{3}}+\sqrt{4-4\sqrt{3}+3}=\frac{1}{2-\sqrt{3}}+\sqrt{{(2-\sqrt{3})}^2} \\ =\frac{1}{2-\sqrt{3}}+2-\sqrt{3}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}+2-\sqrt{3}=\frac{2+\sqrt{3}}{1}+2-\sqrt{3}=4 \end{gathered}$$

$$\begin{gathered} E=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}=\sqrt{\frac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{{\left(2\sqrt{3}\right)}^2-1}}-\sqrt{\frac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{5^2-{\left(2\sqrt{3}\right)}^2}} \\ =\sqrt{\frac{22-11\sqrt{3}}{11}}-\sqrt{\frac{26+13\sqrt{3}}{13}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}} \\ =\sqrt{\frac{4-2\sqrt{3}}{2}}-\sqrt{\frac{4+2\sqrt{3}}{2}}=\frac{1}{\sqrt{2}}\left(\sqrt{{\left(\sqrt{3}-1\right)}^2}-\sqrt{{\left(\sqrt{3}+1\right)}^2}\right) \\ =\frac{1}{\sqrt{2}}\left(\left|\sqrt{3}-1\right|-\sqrt{3}-1\right)=\frac{1}{\sqrt{2}}.(-2)=-\sqrt{2} \end{gathered}$$

$A=(\sqrt{3}+4)\sqrt{19-8\sqrt{3}}=\left(\sqrt{3}+4\right)\sqrt{{\left(4-\sqrt{3}\right)}^2}=\left(\sqrt{3}+4\right)\left(4-\sqrt{3}\right)=16-3=13$

$B={\left(\sqrt{5}\right)}^2-2^2-\frac{\sqrt{{(2-\sqrt{3})}^2}}{\sqrt{3}-2}=5-4-\frac{2-\sqrt{3}}{\sqrt{3}-2}=1-(-1)=2$

$C=\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}-\sqrt{3-2\sqrt{2}}$

Ta có: ${\left(\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}\right)}^2=\frac{14+2\sqrt{44}}{7+2\sqrt{11}}=2\Rightarrow \frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}=\sqrt{2}$$\Rightarrow C=\sqrt{2}-\sqrt{3-2\sqrt{2}}=\sqrt{2}-\sqrt{{\left(\sqrt{2}-1\right)}^2}=\sqrt{2}-\sqrt{2}+1=1$

$$\begin{gathered} D=\sqrt{\frac{4}{{\left(2-\sqrt{5}\right)}^2}}-\sqrt{\frac{4}{{\left(2+\sqrt{5}\right)}^2}}=\frac{\sqrt{2^2}}{\sqrt{{\left(2-\sqrt{5}\right)}^2}}-\frac{\sqrt{2^2}}{\sqrt{{\left(2+\sqrt{5}\right)}^2}} \\ =\frac{2}{\left|2-\sqrt{5}\right|}-\frac{2}{\left|2+\sqrt{5}\right|}=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)} \\ =\frac{2\sqrt{5}+4-2\sqrt{5}+4}{5-4}=8 \end{gathered}$$

$E=\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{1}{\sqrt{15}-1}.\sqrt{\frac{16-2\sqrt{15}}{4}}=\frac{1}{\sqrt{15}-1}.\frac{\sqrt{15}-1}{2}=\frac{1}{2}$

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