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

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

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

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

Bài 13:

1: \(A=-x^2+4x+3\)

\(=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7\le7\)

Dấu '=' xảy ra khi x=2

2: \(B=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\)

Dấu '=' xảy ra khi x=3

Bài 1: 

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

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

b: \(8x^2-48x+6xy-36y\)

\(=8x\left(x-6\right)+6y\left(x-6\right)\)

\(=2\left(x-6\right)\left(4x+3y\right)\)

d: \(a^2-2ab+b^2-4\)

\(=\left(a-b\right)^2-4\)

\(=\left(a-b-2\right)\left(a-b+2\right)\)

a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)

\(\Leftrightarrow5⋮\left(x+1\right)\)

mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)

Vậy...

b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)

Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)

\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)

\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)

Vậy...

c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)

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

Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)

Vậy...

d)\(f\left(x\right)=x^3-3x^2-4x+3=x\left(x^2-3x-4\right)+3=x\left(x+1\right)\left(x-4\right)+3\)

Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)

Vậy...