\(\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\\ =\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{2\sqrt{3}+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\\ =\frac{\sqrt{2\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

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

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

mk ko pit lm tiep dau nha 

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

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

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

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

\(x=-\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(x=-\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(x=-\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(x=-\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

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

\(x=-\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

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

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

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

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

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

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

\(x=-\frac{\sqrt{3}+1}{\sqrt{3}+1}=-1\)

đến đây dễ òi nhé

\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2.\left(2\sqrt{3}\right).1+1}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}}{\sqrt{6}-\sqrt{2}}\)

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

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

cơ bản là lười.ko cần li ke :D

Bài 2

\(P=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}-\sqrt{2}}\)

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

\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

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

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

\(=\frac{\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)

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

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

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

Vậy P là một số nguyên

\(B=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\\ B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ B=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ B=2\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4+\sqrt{15}}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{16-15}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}.\sqrt{5}+\left(\sqrt{5}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)=\left|\sqrt{5}+\sqrt{3}\right|\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)