a) Áp dụng định lý Bézout ( Bê-du ) , dư của \(f\left(x\right)=x^3+x^2-x+a\)cho x + 2 = x - (-2) là \(f\left(-2\right)\)

Để f(x) chia hết cho x + 2 thì f(-2)=0

\(\Rightarrow\left(-2\right)^3+\left(-2\right)^2-\left(-2\right)+a=0\)

\(-8+4+2+a=0\)

\(a-2=0\)

\(a=2\)

Vậy ...

c) \(\frac{n^3+n^2-n+5}{n+2}=\frac{n^3+2n^2-n^2-2n+n+2+3}{n+2}\)nguyên để \(n^3+n^2-n+5⋮n+2\)

\(\Rightarrow\frac{n^2\left(n+2\right)-n\left(n+2\right)+\left(n+2\right)+3}{n+2}\in Z\)

\(\Rightarrow n^2-n+1+\frac{3}{n+2}\in Z\)

\(n^2,n,1\in Z\Rightarrow\frac{3}{n+2}\in Z\)

\(\Rightarrow n+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow n\in\left\{-5;-3;-1;1\right\}\)

Vậy ...

b) Làm tính chia :

\(\Rightarrow-ax+b-a+1=0\)

\(A=\left\{x\in N|x\ge3\right\}\)

\(\Rightarrow A=\left\{3;4;5;6;7;...\right\}\)

\(B=\left\{x\in N|x⋮3,x< 10\right\}\) 

\(\Rightarrow B=\left\{0;3;6;9\right\}\)

a.A={21;24;27;30;33;36;39;42;45;48}

b.B={3}

c.C={0;1;4;9}

a, a bằng {21;24;27;30;33;36;39;42;45;48}

b, b bằng {3}

c, c bằng {0;1;4;9}

a, A=8+12+x = 20 + x

Vì: 20:3 dư 2 => x:3 dư 1 thì A chia hết cho 3

b, A không chia hết cho 3 khi x chia hết cho 3 hoặc x:3 dư 2

a/Ta có: M(x)+N(x) = (2x⁵ - 4x³ + 2x² + 10x - 1) + (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10)

                              = 2x⁵ - 2x⁵ - 4x³ + 4x³ + 2x⁴ + 2x² + x² + 10x + x -1 - 10

                              = 2x⁴ + 3x² + 11x - 11

b/ Ta có: A(x) = N(x)-M(x) = (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10) - (2x⁵ - 4x³ + 2x² + 10x - 1)

                                         = -2x⁵ - 2x⁵ + 2x⁴ + 4x³ + 4x³ + x² - 2x² + x - 10x -10 + 1

                                         = -2x⁵ + 2x⁴ + 8x³ - x² - 9x -9