\(=\frac{2\sqrt{10}-2\sqrt{2}}{2\sqrt{10}-2\sqrt{2}}+\frac{\sqrt{6}\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}.\)

\(=1+\frac{\sqrt{6}}{2\sqrt{2}}=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}.\)

đáp án là \(\sqrt{24}\)anh mình bảo thế nếu sai thì cho mình xin lỗi nha

\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

\(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

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

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

\(=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\sqrt{5}+2}\)

\(=\frac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

Kẻ đường cao \(BH\).

Xét tam giác \(ABH\)vuông tại \(H\): 

\(BH^2=AB^2-AH^2\)

Xét tam giác \(BCH\)vuông tại \(H\):

\(BH^2=BC^2-CH^2=BC^2-\left(AC-AH\right)^2\)

\(=BC^2-AC^2+2AC.AH-AH^2\)

\(\Rightarrow BC^2-AC^2+2AC.AH-AH^2=AB^2-AH^2\)

\(\Leftrightarrow BC^2=AB^2+AC^2-2AC.AH=AB^2+AC^2-2AC.ABcosA\)

\(\frac{x^2-3}{x+\sqrt{3}}\)

\(=\frac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}\)

\(=x-\sqrt{3}\)

#H

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