a) Ta có:
\(A=B\cdot M\Rightarrow M=\dfrac{A}{B}\)
\(\Rightarrow M=\dfrac{7x-6x^3+x^4+5x\left(x+1\right)}{-4+x^2-3x}\)
\(\Rightarrow M=\dfrac{x^4-6x^3+5x^2+7x+5x}{x^2-3x-4}\)
\(\Rightarrow M=\dfrac{x^4-3x^3-4x^2-3x^3+9x^2+12x}{x^2-3x-4}\)
\(\Rightarrow M=\dfrac{x^2\left(x^2-3x-4\right)-3x\left(x^2-3x-4\right)}{x^2-3x-4}\)
\(\Rightarrow M=\dfrac{\left(x^2-3x-4\right)\left(x^2-3x\right)}{x^2-3x-4}\)
\(\Rightarrow M=x^2-3x\)
b) \(M=0\) khi:
\(x^2-3x=0\)
\(\Rightarrow x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a: A=B*M
=>M=A/B
\(\Leftrightarrow M=\dfrac{7x-6x^3+x^4+5x^2+5x}{x^2-3x-4}\)
\(=\dfrac{x^4-6x^3+5x^2+12x}{x^2-3x-4}\)
\(=\dfrac{x^4-3x^3-4x^2-3x^3+9x^2+12x}{x^2-3x-4}\)
=x^2-3x
b: M=0
=>x^2-3x=0
=>x(x-3)=0
=>x=0 hoặc x=3
a) \(A=7x-6x^3+x^4+5x\left(x+1\right)\)
\(=x^4-6x^3+5x^2+12x\)
\(=x\left(x^3-6x^2+5x+12\right)\)
\(=x\left(x^3+x^2-7x^2-7x+12x+12\right)\)
\(=x\left[\left(x^3+x^2\right)-\left(7x^2+7x\right)+\left(12x+12\right)\right]\)
\(=x\left[x^2\left(x+1\right)-7x\left(x+1\right)+12\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(x^2-7x+12\right)\)
\(=x\left(x+1\right)\left(x^2-3x-4x+12\right)\)
\(=x\left(x+1\right)\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\)
\(=x\left(x+1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]\)
\(=x\left(x+1\right)\left(x-3\right)\left(x-4\right)\)
\(B=-4+x^2-3x\)
\(=x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=\left(x^2+x\right)-\left(4x+4\right)\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)
\(A=B.M\Rightarrow M=\dfrac{A}{B}\)
\(=\dfrac{x\left(x+1\right)\left(x-3\right)\left(x-4\right)}{\left(x+1\right)\left(x-4\right)}\)
\(=x\left(x-3\right)\)
b) Cho \(M=0\)
\(\Rightarrow x\left(x-3\right)=0\)
\(\Rightarrow x=0\) hoặc \(x-3=0\)
*) \(x-3=0\)
\(x=3\)
Vậy \(x=0;x=3\)
`#\text {<3 08}`
`a)`
`A = B*M`
`=> M = A \div B`
`A = 7x - 6x^3 + x^4 + 5x(x + 1)`
`= x^4 - 6x^3 + 5x^2 + 5x + 7x`
`= x^4 - 6x^3 + 5x^2 + 12x`
_
`M = A \div B`
`=> M = (x^4 - 6x^3 + 5x^2 + 12x) \div (x^2 - 3x - 4)`
`=> M = x^2 - 3x` (nếu cần phép chia chi tiết thì ib hoặc bl dưới bài làm của mình)
`b)`
`M = x^2 - 3x = 0`
`<=> x*x - 3x = 0`
`<=> x(x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x \in {0; 3}.`
