$b^2+2a\sqrt{c};\:a=5;\:b=7;\:c=9$
$\frac{\left(\left(x\cos\left(t\right)+h\right)-h\right)}{x^2}$
$\left(+93\right)\cdot\left(+98\right)$
$\frac{2x-1}{6}+\frac{x+1}{3}=\frac{3x+2}{4}+\frac{1-x\:}{2}$
$\left(6x^3-3x\right)^2$
$\left(\frac{d^3}{9w^5}\right)^{-2}$
$3x+4+\frac{x}{4}<\frac{5x}{2}+2$
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