$\lim_{x\to\infty}\frac{8x^3-3x^2+1}{4x^3+5x-7}$
$6\left(-a\sqrt{y+x}\right)^2\cdot\left(-b^3\sqrt[4]{z}\right)^2$
$\left(a+5\right)\left(a-5\right)-3\left(a+2\right)\left(a-2\right)+5\left(a+4\right)$
$\frac{3}{4}\:m^3\:y,\:m=4,\:y=11$
$3y^2+5y-2$
$8x+4=6x+6$
$8xe^3x$
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