2x^3+3x^2-x+a x^2+x-1 2x+1 2x^3+x^2 - - 2x^2-x+a 2x^2+x -2x+a -2x-1 - a+1

Để \(A\left(x\right)⋮B\left(x\right)\Leftrightarrow a+1=0\)

                              \(\Leftrightarrow a=-1\)

Vậy ...

2x^3+3x^2-x+a x^2+x-1 2x+1 2x^3+x^2 - - 2x^2-x+a 2x^2+x -2x+a -2x-1 - a+12x^3+3x^2-x+a x^2+x-1 2x+1 2x^3+x^2 - - 2x^2-x+a 2x^2+x -2x+a -2x-1 - a+1

Để \(A \left(\right. x \left.\right) B \left(\right. x \left.\right) \Leftrightarrow a + 1 = 0\)

                              \(\Leftrightarrow a = - 1\)

a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2

1)

a) \(-7x\left(3x-2\right)\)

\(=-21x^2+14x\)

b) \(87^2+26.87+13^2\)

\(=87^2+2.87.13+13^2\)

\(=\left(87+13\right)^2\)

\(=100^2\)

\(=10000\)

2)

a) \(x^2-25\)

\(=x^2-5^2\)

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

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy..........

3)

a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)

Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)

b)

Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)

Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)

c) Cách 1:

x^4+3x^3-x^2+ax+b x^2+2x-3 x^2+x x^4+2x^3-3x^2 - x^3+2x^2+ax+b x^3+2x^2-3x - (a+3)x+b

Để \(P\left(x\right)⋮Q\left(x\right)\)

\(\Leftrightarrow\left(a+3\right)x+b=0\)

\(\Leftrightarrow\hept{\begin{cases}a+3=0\\b=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=-3\\b=0\end{cases}}\)

Vậy a=-3 và b=0 để \(P\left(x\right)⋮Q\left(x\right)\)

a) 

  2n^2-n+2 2n+1 n-1 2x^2+n - -2n+2 -2n-1 - 3

Để \(2n^2-n+2⋮2n+1\)

\(\Leftrightarrow3⋮2n+1\)

\(\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow n\in\left\{0;1;-2;-1\right\}\)

Vậy \(n\in\left\{0;1;-2;-1\right\}\)để \(2n^2-n+2⋮2n+1\)

Có A=\(\left(x^3+x^2-3x\right)+\left(-2x^2-2x+a+2\right)=-x\left(-x^2-x+3\right)-2x^2-2x+a+2⋮-x^2-x+3\)

\(\Rightarrow C=-2x^2-2x+a+2⋮B\). Chỉ có thể C=\(2\left(-x^2-x+3\right)\Rightarrow a+2=6\Rightarrow a=4\)

\(A=\left(2x^3+3x^2+4x\right)+\left(-10x^2-15x+a-8\right)=x\left(2x^2+3x+4\right)+\left(-10x^2-15x+a-8\right)⋮2x^2+3x+4\)\(\Rightarrow C=-10x^2-15x+a-8⋮2x^2+3x+4\)

Chỉ có thể C=\(-5\left(2x^2+3x+4\right)\) \(\Rightarrow a-8=-20\Rightarrow a=-12\)

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

b: Để A chia hết cho B thì \(x+1\in\left\{1;-1;3;-3\right\}\)

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

a, Ta có \(Q\left(x\right)=x+1=0\Leftrightarrow x=-1\)

Vậy P(x) chia hết cho Q(x) khi P(x) có nghiệm là -1 hay

\(3\left(-1\right)^3+2\left(-1\right)^2-5\left(-1\right)+m=0\Leftrightarrow m=-4\)

b.. ta có \(Q\left(x\right)=x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy P(x) chia hết cho Q(x) khi P(x) có nghiệm là 1  và 2 hay

\(\hept{\begin{cases}2+a+b+3=0\\2.2^3+a.2^2+b.2+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=-5\\4a+2b=-19\end{cases}\Leftrightarrow}\hept{\begin{cases}a=-\frac{9}{2}\\b=-\frac{1}{2}\end{cases}}\)