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 \(n\in\left\{3;1;5;-1\right\}\)

Câu b đề thiếu rồi bạn

a: \(\Leftrightarrow2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x+2+a-2⋮x^2-x+1\)

=>a=2

3: \(\Leftrightarrow a-15=0\)

hay a=15

Đặt \(f\left(x\right)=2x^3-3x^2+x+a\)

Ta có: phép chia \(f\left(x\right)\) cho \(x+2\) có dư là \(R=f\left(-2\right)\)

\(\Rightarrow f\left(-2\right)=2.\left(-2\right)^3-3.\left(-2\right)^2+\left(-2\right)+a\)

\(f\left(-2\right)=2.\left(-8\right)-3.4-2+a\)

\(f\left(-2\right)=-16-12-2+a\)

\(f\left(-2\right)=-20+a\)

Để \(f\left(x\right)\) chia hết cho \(x+2\) thì  \(R=0\) hay \(f\left(-2\right)=0\)

\(\Rightarrow-20+a=0\Leftrightarrow a=20\)

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

a: \(\Leftrightarrow2x^2+8x+\left(a-8\right)x+4\left(a-8\right)-4a+28⋮x+4\)

hay a=7

a) đề  x³+x²-x +a chia hét cho (x-1)² ?

x³+x²-x +a=x(x²-2x+1)+3(x²-2x+1)+4x-3+a đề sai nhé

b)A(2)=0=> 8-12+10+m=0  => m=6

c)2n²-n+2=2n(n+1)-3(n+1) +5 chia het cho n+1 khi n+1 là ước của 5

n+1=-1;1;-5;5

n=-2;0;-6;4

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...