Question

[Lil Học Giỏi]

Tìm x , biết :

1, x³ - 2x² - 2x = 0

2, 2x³ - 5x² + 8x - 3 = 0

3, 3x³ - 14x² + 4x + 3 = 0

4, x³ + 6x² - 5x = 0

Answer 1

[Nguyễn Việt Lâm]

a/ \(x\left(x^2-2x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\pm\sqrt{3}\\\end{matrix}\right.\)

b/ \(\Leftrightarrow2x^3-4x^2+6x-x^2+2x-3=0\)

\(\Leftrightarrow2x\left(x^2-2x+3\right)-\left(x^2-2x+3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-2x+3\right)=0\)

c/ \(\Leftrightarrow3x^3-15x^2+9x+x^2-5x+3=0\)

\(\Leftrightarrow3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x^2-5x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=\frac{5\pm\sqrt{13}}{2}\end{matrix}\right.\)

d/ \(x\left(x^2+6x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\pm\sqrt{14}\end{matrix}\right.\)