Question

P(x)=4x³-3x²+4x-4

Q(x)=4x³-x²-3x-4

tìm x để P(x)=Q(x)

Answer 1

\(P\left(x\right)=Q\left(x\right)\Leftrightarrow4x^3-3x^2+4x-4=4x^3-x^2-3x-4\)

<=>\(4x^3-3x^2+4x-4-4x^3+x^2+3x+4=0\)

<=>\(-2x^2+7x=0\Leftrightarrow x\left(-2x+7\right)=0\)<=>x=0 hoặc -2x+7=0

<=>x=0 hoặc x=3,5

Vậy P(x)=Q(x) khi x=0 hoặc x=3,5

Answer 2

Để P(x) = Q(x) <=>\(4x^3-3x^2+4x-4\) = \(4x^3-x^2-3x-4\)

=> \(4x^3-4x^3-3x^2+x^2+4x+3x\) = -4+4 = 0

=>\(-2x^2+7x\) = 0

=>x(-2x+7) = 0

=>\(\left[{}\begin{matrix}x=0\\-2x+7=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=0\\-2x=-7\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{-7}{-2}=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x \(\in\left\{o;\dfrac{7}{2}\right\}\) để P(x)=Q(x)