Câu hỏi của Ngô Ngọc Tú Quỳnh

Question

$P\left(x\right)=2x^4+2x^3-3x^2+x+6$

$Q\left(x\right)=x^4-x^3-x^2+2x+1$

Tính P(x) - 2Q(x)

Mặc Chinh Vũ · Accepted Answer

$-\text{ Có: }Q\left(x\right)=x^4-x^3-x^2+2x+1$

$\Rightarrow2.Q\left(x\right)=2x^4-2x^3-2x^2+4x+2$

$-\text{ Theo chứng minh: }\left\{{}\begin{matrix}P\left(x\right)=2x^4+2x^3-3x^2+x+6\2.Q\left(x\right)=2x^4-2x^3-2x^2+4x+2\end{matrix}\right.$

$\Rightarrow P\left(x\right)-2Q\left(x\right)=\left(2x^4+2x^3-3x^2+x+6\right)-\left(2x^4-2x^3-2x^2+4x+2\right)$

$=2x^4+2x^3-3x^2+x+6-2x^4+2x^3+x^2-2x-1$

$=\left(2x^4-2x^4\right)+\left(2x^3+2x^3\right)+\left(-3x^2+x^2\right)+\left(x-2x\right)+\left(6-1\right)$

$4x^3+\left(-2x^2\right)+\left(-x\right)+5$Câu trả lời: $-\text{ Có: }Q\left(x\right)=x^4-x^3-x^2+2x+1$

$\Rightarrow2.Q\left(x\right)=2x^4-2x^3-2x^2+4x+2$

$-\text{ Theo chứng minh: }\left\{{}\begin{matrix}P\left(x\right)=2x^4+2x^3-3x^2+x+6\2.Q\left(x\right)=2x^4-2x^3-2x^2+4x+2\end{matrix}\right.$

$\Rightarrow P\left(x\right)-2Q\left(x\right)=\left(2x^4+2x^3-3x^2+x+6\right)-\left(2x^4-2x^3-2x^2+4x+2\right)$

$=2x^4+2x^3-3x^2+x+6-2x^4+2x^3+x^2-2x-1$

$=\left(2x^4-2x^4\right)+\left(2x^3+2x^3\right)+\left(-3x^2+x^2\right)+\left(x-2x\right)+\left(6-1\right)$

$4x^3+\left(-2x^2\right)+\left(-x\right)+5$

Violympic toán 7