$\frac{z}{7}>\left(\frac{z}{3}\right)+3$
$3a+4+b+2a+5+b$
$\frac{\left(1+x\right)^6}{x^3}$
$\frac{dm}{dt}=\frac{3}{4}-\frac{3m}{100}$
$\left(2a^2+6bc^3\right)\left(2a^2+bc^3\right)$
$\frac{3}{4}x\cdot6x^2\cdot\left(-2x^3\right)$
$2\left(x-5\right)\le5x+8$
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