Question

a) 3x² - 9x + ( x - 3 ) = 0

b) x³ - 9x - 2x² + 18 = 0

Answer 1

\(3x.\left(x-3\right)+\left(x-3\right)=0\)

\(\left(3x+1\right).\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{3}\end{cases}}\)

vậy \(x=3,x=-\frac{1}{3}\)

\(b,x^3-9x-2x^2+18=0\)

\(x.\left(x^2-9\right)-2.\left(x^2-9\right)=0\)

\(\left(x-2\right).\left(x^2-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x^2=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3,x=-3\end{cases}}\)

vậy \(x=2,x=3,x=-3\)