Để `A(x)\vdotsB(x)` thì `2x^3-x^2+2x^2-x-3x+3/2+m-3/2\vdots2x-1`

`<=>m-3/2=0`

`<=>m=3/2` 

Vậy `m=3/2`

\(2x^3+x^2-4x+m⋮2x-1\)

\(\Leftrightarrow2x^3-x^2+2x^2-x-3x+\dfrac{3}{2}+m-\dfrac{3}{2}⋮2x-1\)

\(\Leftrightarrow m=\dfrac{3}{2}\)

\(A\left(x\right)⋮B\left(x\right)\)

\(\Leftrightarrow2x^3-x^2+2x^2-x-3x+\dfrac{3}{2}+m-\dfrac{3}{2}⋮2x-1\)

\(\Leftrightarrow m=\dfrac{3}{2}\)

a: =>2x^3-4x^2-3x^2+6x+4x-8+a+8 chia hết cho x-2

=>a+8=0

=>a=-8

b: =>2x^3+x^2-x^2-0,5x-0,5x+0,25+m-0,25 chia hết cho 2x+1

=>m-0,25=0

=>m=0,25

\(\Leftrightarrow1-m=0\)

hay m=1

b: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{0;-1;1;-2\right\}\)

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Để a(x) chia hết cho 2x-1 thì \(2x^3-x^2+2x^2-x-3x+\dfrac{3}{2}+m-\dfrac{3}{2}⋮2x-1\)

\(\Leftrightarrow m-\dfrac{3}{2}=0\)

hay \(m=\dfrac{3}{2}\)

a: \(\Leftrightarrow x^3+2x^2-3x^2-6x+5x+10+a-10⋮x+2\)

=>a-10=0

=>a=10

b: \(\Leftrightarrow x^3+x^2+x+\left(a-1\right)x^2+\left(a-1\right)x+a-1+\left(2-a\right)x+b-a+1⋮x^2+x+1\)

=>2-a=0 và b-a+1=0

=>a=2; b=a-1=2-1=1