Bài 1: 

a: \(=\dfrac{2x^4-8x^3+2x^2+2x^3-8x^2+2x+18x^2-72x+18+56x-15}{x^2-4x+1}\)

\(=2x^2+2x+18+\dfrac{56x-15}{x^2-4x+1}\)

Bài 1:

Ta có: \(5x^3-3x^2+2x+a⋮x+1\)

\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)

\(\Leftrightarrow a-10=0\)

hay a=10

f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+5=t\), ta có :

\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)

Thay và ta có :

\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)

\(=2x^2-3x+1\)

a)\(f\left(x\right)=5x^3-9x^2+2x+m=5x^2\left(x+2\right)-19x\left(x+2\right)+40\left(x+2\right)-80+m=\left(x+2\right)\left(5x^2-19x+40\right)+m-80\)

Để \(f\left(x\right)⋮g\left(x\right)\) thì \(m-80=0\Leftrightarrow m=80\)

b) \(f\left(x\right)=\left(x+2\right)\left(5x^2-19x+40\right)+m-80\)

Để f(x) chia g(x) có số dư bằng 3 thì \(m-80=3\Leftrightarrow m=83\)

Áp dụng định lý Bơ-du ta có:

f(x) ⋮ (2x+5) ⇔ f(\(\dfrac{-5}{2}\))=0 (Kiến thức nâng cao lớp 8)

                    ⇔ (\(\dfrac{-5}{2}\))⁴+(\(\dfrac{-5}{2}\))³+12.(\(\dfrac{-5}{2}\))²-m=0

                    ⇔ \(\dfrac{1575}{16}\)-m=0

                    ⇔ m= \(\dfrac{1575}{16}\)

Vậy m=\(\dfrac{1575}{16}\) thì x⁴+x³+12x²-m chia hết cho 2x+5

Bài 3:

Ta có: \(2n^2+n-7⋮n-2\)

\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{3;1;5;-1\right\}\)

a: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)

\(=2x^2-3x+1\)