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

\(=2\sqrt{3}-2+10+5\sqrt{3}-\frac{2\sqrt{3}\left(\sqrt{3}+1\right)}{2}=7\sqrt{3}+8-3-\sqrt{3}=5+6\sqrt{3}\)

a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)

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

Đầu hàng thôi,tôi ko bt ♤♡◇♧><

thực hiện phép tính nha cám ơn m.ng

a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)

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

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

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)

\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)

c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)

d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)

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

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

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

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

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

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

\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)

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

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)

\(=\dfrac{\sqrt{1\cdot4}}{2}\)

\(=\dfrac{2}{2}\)

\(=1\)