$\lim_{x\to0}\frac{ln\left(\tan\left(x\right)^2\right)}{lnx}$
$\left(x^2+9\right)^8=0$
$-3m\cdot5m^4-3m^3+6m-3$
$-8+6xy-19m+12xy+20m+4+23-5xy$
$\frac{d}{dx}\frac{9x-1}{5}$
$1^3+2^3$
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