\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Leftrightarrow5x-10x^2-3x^2-54x=0\)
\(\Leftrightarrow-13x^2-49x=0\)
\(\Leftrightarrow-x\left(13x-49\right)=0\)
\(\Rightarrow\hept{\begin{cases}-x=0\Leftrightarrow x=0\\13x-49=0\Leftrightarrow x=\frac{49}{13}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{49}{13}\)
5 (1-2x) - 3x(x+18) = 0
5x - 10x2 - 3x 2- 54x = 0
- 13x2 - 49x = 0
(-13x - 49)x = 0
TH1
:x = 0
TH2:
-13x - 49 = 0
-13x = 49
x = -49/13
vậy x = 0 và x = -49/13