Question

(\(\frac{2}{3}\)+x).(\(\frac{1}{5}\)-2x)=0. Ai giúp em với ạ, em cảm ơn nhìu :3

Answer 1

\(\left(\frac{2}{3}+x\right)\left(\frac{1}{5}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}+x=0\\\frac{1}{5}-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\-2x=-\frac{1}{5}\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-\frac{2}{3}\\x=\frac{1}{10}\end{cases}}\)

Vậy:.......

#H

Answer 2

\(\left(\frac{2}{3}+x\right)\left(\frac{1}{5}-2x\right)=0\)

\(\Rightarrow\frac{2}{3}+x=0\)hoặc\(\frac{1}{5}-2x=0\)

\(\Rightarrow x=-\frac{2}{3}\)   hoặc\(2x=\frac{1}{5}\)

\(\Rightarrow x=-\frac{2}{3}\)   hoặc\(x=\frac{1}{10}\)