a) \(4x^2-4x-15=0\)
<=> \(\left(2x-5\right)\left(2x+3\right)=0\)
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
Vậy....
b) \(x^3-6x^2-x+30=0\)
<=> \(\left(x-5\right)\left(x-3\right)\left(x+2\right)=0\)
tự giải tiếp