\(\sqrt{7-4\sqrt{3}}+\frac{1}{2-\sqrt{3}}.\)

\(\Leftrightarrow\sqrt{4-2.2.\sqrt{3}+3}+\frac{1}{2-\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(2+\sqrt{3}\right)^2}+\frac{1}{2-\sqrt{3}}\)

\(\Leftrightarrow\left(2+\sqrt{3}\right)+\frac{1}{2-\sqrt{3}}\)

\(\Leftrightarrow\frac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{2-\sqrt{3}}+\frac{1}{2-\sqrt{3}}\)

\(\Leftrightarrow\frac{4-3+1}{2-\sqrt{3}}=\frac{2}{2-\sqrt{3}}\)

\(\sqrt{7-4\sqrt{3}}+\frac{1}{2-\sqrt{3}}\)

\(=\sqrt{\sqrt{3}^2-2.2.\sqrt{3}+2^2}+\frac{2}{2\left(2-\sqrt{3}\right)}\)

\(=\sqrt{\left(\sqrt{3}-2\right)^2}-\frac{1}{\sqrt{3}-2}\)

\(=\sqrt{3}-2-\frac{1}{\sqrt{3}-2}=\frac{\left(\sqrt{3}-2\right)^2-1}{\sqrt{3}-2}\)

\(=\frac{7-4\sqrt{3}-1}{\sqrt{3}-2}=\frac{6-4\sqrt{3}}{\sqrt{3}-2}\)

ĐK: Tự tìm 

Đặt \(\left(x+2\right)=u;\left(y+6\right)=v\)

phương trình (2) <=> \(2\sqrt{\left(x+2\right)\left(3x+6-6-y\right)}=y+6\)

=>  \(2\sqrt{u\left(3u+v\right)}=v\)

<=> \(\hept{\begin{cases}4u\left(3u+v\right)=v^2\\v\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}12u^2+4uv=v^2\left(1'\right)\\v\ge0\left(2'\right)\end{cases}}\)

(1') <=> \(v^2-4uv-12u^2=0\)

\(\Delta_u=4u^2+12u^2=16u^2\)

=> \(\orbr{\begin{cases}v=2u+4u=6u\\v=2u-4u=-2u\end{cases}}\)

Với v = 6u ta có:  y + 6 = 6 ( x + 2 ) <=> y = 6x + 6  thế vào phương trình ban đầu => tìm x, y

 Với  v = - 2u  ta có:.... 

Thế nhé! Em làm tiếp! 

\(\sqrt{14}-6\sqrt{5}-\sqrt{21}-8\sqrt{5}.\)

\(=\left(-6\sqrt{5}-8\sqrt{5}\right)+\left(\sqrt{14}-\sqrt{21}\right)\)

\(=-14\sqrt{5}+\sqrt{7}\left(\sqrt{2}-\sqrt{3}\right)\)