giải các pt : a; x^3 – 3x^2 +x+5 = 0 b; 2x^3 +x^2 -4x +1=0 30/07/2021 Bởi Margaret giải các pt : a; x^3 – 3x^2 +x+5 = 0 b; 2x^3 +x^2 -4x +1=0
a) $x^3-3x^2+x+5=0$ $↔ x^3+x^2-4x^2-4x+5x+5=0$ $↔ x^2(x+1)-4x(x+1)+5(x+1)=0$ $↔ (x+1)(x^2-4x+5)=0$ $↔ \left[ \begin{array}{l}x+1=0\\x^2-4x+5=0\end{array} \right.$ $→ x=-1$ b) $2x^3+x^2-4x+1=0$ $↔ 2x^3-2x^2+3x^2-3x-x+1=0$ $↔ 2x^2(x-1)+3x(x-1)-(x-1)=0$ $↔ (x-1)(2x^2+3x-1)=0$ $↔ \left[ \begin{array}{l}x-1=0\\2x^2+3x-1=0\end{array} \right.$ $↔ \left[ \begin{array}{l}x=1\\x=\dfrac{-3+\sqrt[]{17}}{4}\\x=\dfrac{-3-\sqrt[]{17}}{4}\end{array} \right.$ Bình luận
a) $x^3-3x^2+x+5=0$
$↔ x^3+x^2-4x^2-4x+5x+5=0$
$↔ x^2(x+1)-4x(x+1)+5(x+1)=0$
$↔ (x+1)(x^2-4x+5)=0$
$↔ \left[ \begin{array}{l}x+1=0\\x^2-4x+5=0\end{array} \right.$
$→ x=-1$
b) $2x^3+x^2-4x+1=0$
$↔ 2x^3-2x^2+3x^2-3x-x+1=0$
$↔ 2x^2(x-1)+3x(x-1)-(x-1)=0$
$↔ (x-1)(2x^2+3x-1)=0$
$↔ \left[ \begin{array}{l}x-1=0\\2x^2+3x-1=0\end{array} \right.$
$↔ \left[ \begin{array}{l}x=1\\x=\dfrac{-3+\sqrt[]{17}}{4}\\x=\dfrac{-3-\sqrt[]{17}}{4}\end{array} \right.$