\(\left(2x-\frac{3}{7}\right).\left(2x^2+18\right)=0\)
* \(2x-\frac{3}{7}=0\) *\(2x^2+18=0\)
\(2x=\frac{3}{7}\) \(2x^2=-18\)
\(x=\frac{3}{7}:2\) vô lí vì 2x2 \(\ne\)-18
\(x=\frac{3}{14}\)
vậy x=\(\frac{3}{14}\)
làm lại TH2 nha:
2x2+18=0
2x2=0+18
2x2=-18
x2=-18:2
x2=-9
vô lí vì x2 ko bằng -9
\(\left(2x-\frac{3}{7}\right).\left(2x^2+18\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{3}{7}=0\\2x^2+18=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{3}{7}\\2x^2=-18\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{7}\cdot\frac{1}{2}=\frac{3}{14}\\x^2=-9vôli\end{cases}}\)
\(\Rightarrow x=\frac{3}{14}\)
Vậy ....
