a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)

\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)

\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)

b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)

\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)

\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)

c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)

 Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề hơn nhé. 

a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

1.12=12

2.56=112

1 . 12 = 12

2 . 56 = 112

k tớ nhé