\(\Leftrightarrow\dfrac{3\left(x+7\right)}{15}+\dfrac{5\left(4x+5\right)}{15}\ge0\)
\(\Leftrightarrow3\left(x+7\right)+5\left(4x+5\right)\ge0\)
\(\Leftrightarrow23x+46\ge0\)
\(\Leftrightarrow23x\ge-46\)
\(\Leftrightarrow x\ge-2\)
Lời giải:
$\frac{x+7}{5}+\frac{4x+5}{3}\geq 0$
$\Leftrightarrow \frac{x}{5}+\frac{4x}{3}+\frac{7}{5}+\frac{5}{3}\geq 0$
$\Leftrightarrow \frac{23}{15}x+\frac{46}{15}\geq 0$
$\Leftrightarrow 23x+46\geq 0$
$\Leftrightarrow 23x\geq -46$
$\Leftrightarrow x\geq -2$