\(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}15x-35=0\\5x+3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-3}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{-3}{5}\right\}\)
3x(25x + 15) - 35(5x + 3) = 0
<=> 15x(5x + 3) - 35(5x + 3) = 0
<=> (5x + 3)(15x - 35) = 0
<=> 5(5x + 3)(3x - 7) = 0
<=> 5x + 3 = 0 hay 3x - 7 = 0 (vì 5 \(\ne\)0)
<=> 5x = -3 I <=> 3x = 7
<=> x =\(\frac{-3}{5}\)I <=> x = \(\frac{7}{3}\)
Vậy S = {\(\frac{-3}{5}\); \(\frac{7}{3}\)}
