\(\Leftrightarrow\dfrac{6\left(4x+1\right)-4\left(5x+2\right)}{24}< \dfrac{7\left(x+1\right)}{24}\)
\(\Leftrightarrow6\left(4x+1\right)-4\left(5x+2\right)< 7\left(x+1\right)\)
\(\Leftrightarrow24x+6-20x-8< 7x+7\)
\(\Leftrightarrow-3x< 9\)
\(\Leftrightarrow x>-3\)
Vậy \(S=\left\{x|x>-3\right\}\)
Đúng 2
Bình luận (5)