\(\left(x-\frac{3}{5}\right)\left(x+\frac{2}{7}\right)< 0\)
=> \(x-\frac{3}{5}\)và \(x+\frac{2}{7}\)là 2 số trái dấu
Mà \(x-\frac{3}{5}< x+\frac{2}{7}\)\(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}< 0\\x+\frac{2}{7}>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< \frac{3}{5}\\x>\frac{-2}{7}\end{cases}}\)\(\Rightarrow\frac{-2}{7}< x< \frac{3}{5}\)
Vậy ...
\(\left(x-\frac{3}{5}\right)\left(x+\frac{2}{7}\right)< 0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{5}< 0\\x+\frac{2}{7}< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x< \frac{3}{5}\\x< -\frac{2}{7}\end{cases}}\)
Để tích đó <0 \(\Leftrightarrow x-\frac{3}{5}vàx+\frac{2}{7}\)cùng dấu
\(+\)TH1:
\(x-\frac{3}{5}>0\)và\(x+\frac{2}{7}>0\)\(\Rightarrow x>\frac{3}{5}\)và \(x< -\frac{2}{7}\)\(\Rightarrow\)loại
+TH2
\(x-\frac{3}{5}< 0\\ x+\frac{2}{7}< 0\\\)\(\Rightarrow x< \frac{3}{5}\\ x>-\frac{2}{7}\)\(\Rightarrow-\frac{2}{7}< x< \frac{3}{5}\)
Vậy \(-\frac{2}{7}< x< \frac{3}{5}\)
