Question

a.4x^3-4x^2+x=0

b.x.(x-3)+12-4x=0

c.x^3+3x^2+3x-7=0

*tìm x*

 

Answer 1

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)