$\frac{\tan\left(x\right)+\cos\left(x\right)}{\sec\left(x\right)+\cot\left(x\right)}$
$\left(-2m^3-m^2+3\right)$
$x^2+x+18$
$10-\left|2-\left(6-10\right)\right|$
$\frac{x^2-5x+2}{x+6}$
$6+2f-7$
$8x^3+14x^2$
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