Bài 2:

x^3+6x^2+12x+m chia hết cho x+2

=>x^3+2x^2+4x^2+8x+4x+8+m-8 chia hết cho x+2

=>m-8=0

=>m=8

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

a) \(x\left(x-2\right)-x+2=0\)

\(x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(x-1\right)\left(x-2\right)=0\)

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

\(a,\Leftrightarrow2x^2-10x-2x^2-x=-11\\ \Leftrightarrow-11x=-11\Leftrightarrow x=1\\ b,\Leftrightarrow x\left(x^2-6x+9\right)=0\\ \Leftrightarrow x\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-2018\right)-2017\left(x-2018\right)=0\\ \Leftrightarrow\left(x-2017\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2018\end{matrix}\right.\)