5x(1-2x)-3x(x+18)=0
o) TH1 :
5x(1-2x)=0
=> x(1-2x)=0
=> x-2x2=0
=> x=0
o) TH2 :
3x(x+18)=0
=> x(x+18)=0
=> x+18=0
=> x=-18
Vậy x=0 hoặc x=-18.
\(5x\left(2-2x\right)-3x\left(x+18\right)=0\)
\(\Leftrightarrow10x-10x^2-3x^2-54x=0\Leftrightarrow-44x-13x^2=0\)
\(\Leftrightarrow x\left(-13x-44\right)=0\)
TH1 : x = 0 ; TH2 : x = -44/13
5x( 1 - 2x ) - 3x( x + 18 ) = 0
<=> 5x - 10x2 - 3x2 - 54x = 0
<=> -13x2 - 49x = 0
<=> -x( 13x + 49 ) = 0
<=> \(\orbr{\begin{cases}-x=0\\13x+49=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{49}{13}\end{cases}}\)