Question

\(cho\) \(đa\) \(thức\): \(P\left(x\right)=ax^3+bx^2+cx+d\) \(\left(a\ne0\right)\)

\(bt:\) \(P\left(1\right)=100\) ; \(P\left(-1\right)=50\) ; \(P\left(2\right)=120\) . \(tính\) \(P\left(3\right)\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c+d=100\\a-b+c-d=-50\\8a+4b+2c+d=120\\27a+9b+3c+d=P\left(3\right)\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\\\left(4\right)\end{matrix}\)

(1)+(2) \(\Leftrightarrow2\left(a+c\right)=50\Rightarrow c=25-a\)

(1)-(2) \(\Leftrightarrow2\left(b+d\right)=150\Rightarrow b=75-d\)

thế vào (3)<=> \(8a+4\left(75-d\right)+2\left(25-a\right)+d=120\)

\(\Leftrightarrow6a-3d=230\Rightarrow d=2a+\dfrac{230}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=25-a\\b=-2a-\dfrac{5}{3}\\d=2a+\dfrac{230}{3}\end{matrix}\right.\)

\(P\left(3\right)=27a-9\left(2a+\dfrac{5}{3}\right)+3\left(25-a\right)+2a+\dfrac{230}{3}\)

\(\left\{{}\begin{matrix}\forall a\in R;a\ne0\\P\left(3\right)=8a+\dfrac{410}{3}\end{matrix}\right.\)

Answer 2

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