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

\(A=\sqrt{5}.\left(\sqrt{20}-3\right)+\sqrt{45}.\)

\(=\sqrt{5}.\left(\sqrt{4.5}-3\right)+\sqrt{9.5}\)

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

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

\(=2.\sqrt{5}^2=2.5=10\)

a: \(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}=14\sqrt{3}-2\sqrt{57}\)

b: \(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)

b: \(=8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)

c: \(=2\sqrt{21}\)

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

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

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