\(\frac{3}{x+3}>\frac{1}{x}>\frac{4}{x+7}\)
\(\Leftrightarrow\frac{3.4}{4\left(x+3\right)}>\frac{12}{12.x}>\frac{3.4}{\left(x+7\right).3}\)
\(\Leftrightarrow\frac{12}{4x+12}>\frac{12}{12x}>\frac{12}{3x+21}\Leftrightarrow4x+12<12x<3x+21\)
\(\Leftrightarrow\int^{4x+12<12x}_{12x<3x+21}\Leftrightarrow\int^{12<8x}_{8x<21}\)
\(\Leftrightarrow12<8x<21\Leftrightarrow8x\in\left\{16\right\}\Leftrightarrow x=2\)
Vậy x=2