$\frac{\left(\sin\left(x\right)+\cos\left(x\right)\right)^2-1}{\sin\left(x\right)}$
$\log\left(3+x\right)=1-\log\left(4-x\right)$
$\frac{x^3+3x^2+9}{-3}$
$3a+5\cdot\left(2x-1\right)$
$-6-\left(-5\right)$
$4x^2+20x+9$
$2x+3+7x=-24$
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