Question

a) ( x - 2 )³ + 6 ( x + 1 )² - x³ + 12 = 0

b) ( x + 2 ) ( x² - 2x + 4 ) - x ( x - 3 ) ( x + 3 ) = 26

c) 3x² + 6x = 0 

d) 3x³  - x = 0

Answer 1

a: \(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

=>24x+10=0

hay x=-5/12

b: \(\Leftrightarrow x^3+8-x\left(x^2-9\right)=26\)

=>9x=18

hay x=2

c: =>3x(x+2)=0

=>x=0 hoặc x=-2

Answer 2

`d)3x^3 -x=0`

`=>x(3x^2 -1)=0`

\(\Rightarrow\left[{}\begin{matrix}x=0\\3x^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{1}{3}\end{matrix}\right.\)

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

`=>` Vây...