$\frac{x^2+10x+16}{x^2-4}$
$5xy^'+3xy=x\cos\left(2x\right)^{\:\:}+x$
$12x^2+3x+6$
$t^2+16t+64$
$\frac{\left(3x^2+5\right)}{\left(\left(x-1\right)\left(x+1\right)^2\right)}$
$6x\left(2x+1\right)$
$25\left(x-y\right)^2$
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