Question

[Lê Ngọc Chí]

tìm nghiệm của đa thức sau: -4x^3 +4x

Answer 1

[Nguyễn Thị Thương Hoài]

-4\(x^3\) + 4\(x\) = 0

- 4\(x\) ( \(x^2\) - 1) = 0

\(\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

Answer 2

[Duc Nguyen]

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

Áp dụng công thức phương trình bậc 3, ta có:

\(a=-4,b=0,c=4,d=0\)

\(\Rightarrow\Delta=b^2-3ac=0^2-3\cdot-4\cdot4=0+48=48\)

\(\Rightarrow k=\dfrac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^3}}\)

\(\Rightarrow k=\dfrac{9\cdot-4\cdot0\cdot4-2\cdot0^3-27\cdot\left(-4\right)^2\cdot0}{2\sqrt{\left|48\right|^3}}\)

\(\Rightarrow k=\dfrac{0}{2\sqrt{\left|48\right|^3}}=0\)

Vì Δ = 48 > 0 và k = 0 < 1

\(\Rightarrow x_1=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}\right)-b}{3a}\)

\(x_1=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)}{3}\right)-0}{3\cdot-4}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}}{3}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\pi}{6}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{\dfrac{8\sqrt{3}\cdot\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{4\cdot3}{-12}=\dfrac{12}{-12}=-1\)

\(\Rightarrow x_2=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}-\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_2=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{-3\pi}{2}}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{-3\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{-\pi}{2}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}\cdot0}{-12}=0\)

\(\Rightarrow x_3=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}+\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_3=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)+2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}+2\pi}{3}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{5\pi}{2}}{3}\right)}{-12}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{5\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}\cdot\dfrac{-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\sqrt{3}\cdot-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\cdot-3}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{-24}{2}}{-12}\)

\(x_3=\dfrac{-12}{-12}=1\)

Vậy: \(x_1=-1,x_2=0,x_3=1\)