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 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

 Thay  vào biểu thức ta được:

Ta có:

Vậy giá trị biểu thức bằng y – (-x) = x + y.

\(a,B=4x^2+20x+25-9+x^2+14=5x^2+20x+30\\ b,B=5\left(x^2+4x+4\right)+10\\ B=5\left(x+2\right)^2+10\ge10>0,\forall x\)

Do đó B luôn dương với mọi x

đề bài lỗi bn ơi

a) a ≠ 0 ,    a ≠   − 5  

b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2  

c) Thay a = -1 (TMĐK) vào a ta được A = -1

d) Ta có A = 0 Û a = 1 (TMĐK)

a,ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\x^2-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\\x\ne\pm3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

b, \(M=\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}-\dfrac{36}{x^2-9}\)

\(\Rightarrow M=\dfrac{\left(x+3\right)^2-\left(x-3\right)^2-36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{x^2+6x+9-x^2+6x-9-36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{12x-36}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{12\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{12}{x+3}\)

c, Thay x=0 vào M ta có: \(M=\dfrac{12}{x+3}=\dfrac{12}{0+3}=\dfrac{12}{3}=4\)

a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

b: \(M=\dfrac{x^2+6x+9-x^2+6x-9-36}{\left(x-3\right)\left(x+3\right)}=\dfrac{12x-36}{\left(x-3\right)\left(x+3\right)}=\dfrac{12}{x+3}\)

c: Thay x=0 vào M, ta được:

\(M=\dfrac{12}{0+3}=\dfrac{12}{3}=4\)