$\frac{\left(3x+2\right)}{\left(x-1\right)\left(x+4\right)}$
$x^2+16\ge0$
$\frac{3x}{x^2-9}\cdot\:\frac{3}{3-x}$
$-6x^2+7x+4x^2$
$\left(3a^2-5b\right);\:a=1;\:b=-1$
$-a^2b^8-c^4$
$7y-9+18-16y+5$
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