Question

Tìm x

\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

Answer 1

3x - 1/4 = 0 hay x + 1/2 = 0

3x= 1/4 hay x = -1/2

x = 1/12 hay x = -1/2

Answer 2

Ko chép lại đề bài

=> 3x - 1/4 = 0 hoặc x +1/2 = 0

Nếu 3x-1/4 = 0 thì 3x = 0+1/4 => 3x/ 1/4 => x= 1/4 :3 => x= 1/12

Nếu x+1/2 = 0 thì x = 0-1/2 => x= -1/2

Vậy...

Ko chắc nha

Answer 3

\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{1}{2}\end{cases}}}\)

Answer 4

\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)