\(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}=\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}=\dfrac{-4\sqrt{6}}{25-24}=-4\sqrt{6}\)

\(=\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}=\dfrac{-4\sqrt{6}}{25-24}=-4\sqrt{6}\)

\(\left(\dfrac{6-2}{1-3}-\dfrac{5}{5}\right):\dfrac{1}{5-2}\)

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

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

\(=-3.3\)

\(=-9\)

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

=(-3)*3

=-9

cái đó là toán lớp 5 nha bẩm lộn ấ

a, 5/10 - 2/15 = 1/2 - 2/15 = 15/30 - 4/30 = 11/30

b,5/20 - 1/6 = 1/4 - 1/6 = 3/12 - 2/12 = 1/12

c,6/18 - 6/24 = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 

d,5/9 - 3/12 = 5/9 - 1/4 = 20/36 - 9/36 = 11/36 
chắc là đúng á , tại lớp 8 rồi ko nhớ lắm , mà vote mình nhá

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

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

A = 4

Đặt \(x=\sqrt{\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}}>0\)

\(x^2=\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}+2\sqrt{\dfrac{25-24}{25-6}}=\dfrac{74}{19}+\dfrac{2\sqrt{19}}{19}\)

\(\Rightarrow x^2=\dfrac{74+2\sqrt{19}}{19}\Rightarrow x=\sqrt{\dfrac{74+2\sqrt{19}}{19}}\)

Ko thể rút gọn thêm nữa (có thể trục căn thức ở mẫu)

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

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

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

\(=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

\(=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)