Question

a) ( x - 1/5 ) . ( x + 4/7) > 0

b) ( x + 2/3 ) . ( x + 2 ) < 0

Giúp mình với ạ đang cần gấp 

Answer 1

a) Ta có: \(\left(x-\frac{1}{5}\right).\left(x+\frac{4}{7}\right)>0\)

   + \(\hept{\begin{cases}x-\frac{1}{5}>0\\x+\frac{4}{7}>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>\frac{1}{5}\\x>-\frac{4}{7}\end{cases}}\)\(\Rightarrow\)\(x>\frac{1}{5}\)

   + \(\hept{\begin{cases}x-\frac{1}{5}< 0\\x+\frac{4}{7}< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< \frac{1}{5}\\x< -\frac{4}{7}\end{cases}}\)\(\Rightarrow\)\(x< -\frac{4}{7}\)

Vậy \(x>\frac{1}{5}\)hoặc \(x< -\frac{4}{7}\)

Answer 2

b) Ta có: \(\left(x+\frac{2}{3}\right).\left(x+2\right)< 0\)

   + \(\hept{\begin{cases}x+\frac{2}{3}>0\\x+2< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>-\frac{2}{3}\\x< -2\end{cases}}\)\(\Rightarrow\)\(-\frac{2}{3}< x< -2\)( vô lí )

    + \(\hept{\begin{cases}x+\frac{2}{3}< 0\\x+2>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< -\frac{2}{3}\\x>-2\end{cases}}\)\(\Rightarrow\)\(-\frac{2}{3}>x>-2\)

Vậy \(-2< x< -\frac{2}{3}\)

Answer 3

a) Xét hai trường hợp :

1. \(\hept{\begin{cases}x-\frac{1}{5}>0\\x+\frac{4}{7}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{5}\\x>-\frac{4}{7}\end{cases}}\Leftrightarrow x>\frac{1}{5}\)

2. \(\hept{\begin{cases}x-\frac{1}{5}< 0\\x+\frac{4}{7}< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{5}\\x< -\frac{4}{7}\end{cases}}\Leftrightarrow x< -\frac{4}{7}\)

Vậy \(\orbr{\begin{cases}x>\frac{1}{5}\\x< -\frac{4}{7}\end{cases}}\)

b) Xét hai trường hợp :

1. \(\hept{\begin{cases}x+\frac{2}{3}>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-\frac{2}{3}\\x< -2\end{cases}}\)( loại )

2. \(\hept{\begin{cases}x+\frac{2}{3}< 0\\x+2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -\frac{2}{3}\\x>-2\end{cases}}\Leftrightarrow-2< x< -\frac{2}{3}\)

Vậy -2 < x < -2/3