Question

[Nam Review]

tìm x 

7x^3+3x^2-3x+1=0

Answer 1

[Nguyễn Hoàng Minh]

\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\\ \Rightarrow\left(x+1\right)\left(7x^2-4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\7x^2-4x+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Rightarrow7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{4}{49}\right)+\dfrac{3}{7}=0\\ \Rightarrow7\left(x-\dfrac{2}{7}\right)^2+\dfrac{3}{7}=0\left(\text{vô lí}\right)\)

Vậy x=-1

Answer 2

[Minh Bao]

\(x^2\left(7x-3\right)=0\)