\(d,=\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{6}-\sqrt{3}+\sqrt{2}+1\\ e,=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =2+\sqrt{3}-2\sqrt{5}-1-2\sqrt{5}+2=\sqrt{3}-4\sqrt{5}+3\\ \\ f,=2-\sqrt{3}+\sqrt{3}-1+4=5\\ g,=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\sqrt{2}\\ =\sqrt{3}-\sqrt{3}-\sqrt{2}=-\sqrt{2}\)
\(h,\) Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ A^2=4+2\sqrt{1}=6\\ A=\sqrt{6}\)
g: Ta có: \(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\dfrac{2}{\sqrt{2}}\)
\(=\sqrt{3}-\sqrt{3}-\sqrt{2}\)
\(=-\sqrt{2}\)
h: Ta có: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}\)
\(=\sqrt{6}\)
d: ta có: \(\sqrt{7+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{6}+1-\sqrt{3}+\sqrt{2}\)
e: ta có: \(\sqrt{7+4\sqrt{3}}-\sqrt{21+4\sqrt{5}}-\sqrt{24-8\sqrt{5}}\)
\(=2+\sqrt{3}-2\sqrt{5}-1-2\sqrt{5}+2\)
\(=3+\sqrt{3}-4\sqrt{5}\)
