Question

Tìm x, biết:
\(\left(2x+1\right)\times\left(3x-\frac{9}{2}\right)=0\)

Answer 1

\(\left(2x+1\right).\left(3x-\frac{9}{2}\right)=0\)

=> \(\orbr{\begin{cases}2x+1=0\\3x-\frac{9}{2}=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=-1\\3x=\frac{9}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)

KL: \(x\in\left\{\frac{-1}{2};\frac{3}{2}\right\}\)

Answer 2

\(\left(2x+1\right).\left(3x-\frac{9}{2}\right)=0\)

=>\(\orbr{\begin{cases}2x+1=0\\3x-\frac{9}{2}=0\end{cases}}\)

=>\(\orbr{\begin{cases}2x=-1\\3x=\frac{9}{2}\end{cases}}\)

=>\(\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)

Vậy hoặc \(x=\frac{-1}{2}\)hoặc \(x=\frac{3}{2}\)

k mình nha !!!