1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

Bài 1:

a: \(\sqrt{50}+2\sqrt{8}-\dfrac{3}{2}\cdot\sqrt{72}+\sqrt{125}\)

\(=5\sqrt{2}+2\cdot2\sqrt{2}-\dfrac{3}{2}\cdot6\sqrt{2}+\sqrt{125}\)

\(=9\sqrt{2}-9\sqrt{2}+5\sqrt{5}=5\sqrt{5}\)

b: \(\left(3\sqrt{2}-\sqrt{5}\right)^2-\dfrac{9}{\sqrt{5}-\sqrt{2}}\)

\(=18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5-\dfrac{9\left(\sqrt{5}+\sqrt{2}\right)}{5-2}\)

\(=23-6\sqrt{10}-3\left(\sqrt{5}+\sqrt{2}\right)\)

\(=23-6\sqrt{10}-3\sqrt{5}-3\sqrt{2}\)

c: \(5\sqrt{4a}-3\sqrt{25a}+\sqrt{9a}\)

\(=5\cdot2\sqrt{a}-3\cdot5\sqrt{a}+3\sqrt{a}\)

\(=10\sqrt{a}-15\sqrt{a}+3\sqrt{a}=-2\sqrt{a}\)

\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)

\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)

\(=\left(\sqrt{a}+\sqrt{a}\right).\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=2\sqrt{a}.\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=1+\sqrt{a}\)

(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)

b. \(=\left(\dfrac{\sqrt{a}-a+a\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)

\(=\left(\dfrac{2\sqrt{a}}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)

\(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)

\(=1-a\)

a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)

\(=\sqrt{36}-\sqrt{16}=6-4=2\)

b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)

b. \(=\left(\dfrac{2}{a\left(a+1\right)}-\dfrac{2}{a+1}\right):\dfrac{1-a}{a^2+2a+1}\)

\(=\left(\dfrac{2-2a}{a\left(a+1\right)}\right):\dfrac{1-a}{\left(a+1\right)^1}\)

\(=\dfrac{\left(2-2a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}\)

\(=\dfrac{2\left(1-a\right)\left(a+1\right)^2}{a\left(a+1\right)\left(1-a\right)}=\dfrac{2\left(a+1\right)}{a}\)

a.\(=\sqrt{2}.\left(\sqrt{25}-\sqrt{9}\right)=\sqrt{2}.\left(5-3\right)=2\sqrt{2}\)

a: \(=5\sqrt{2}-3\sqrt{2}=2\sqrt{2}\)

b: \(=\dfrac{2-2a}{a\left(a+1\right)}\cdot\dfrac{\left(a+1\right)^2}{1-a}=\dfrac{2a+2}{a}\)

a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)

\(=\dfrac{x^2+5x+100}{5x}\)

b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:

\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)

Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)