$\frac{160x^{-3}y^2}{-40x^2y}$
$sinx-sin3x$
$\int\frac{\frac{1}{3}x+\frac{1}{3}}{\frac{3}{4}+\left(x-\frac{1}{2}\right)^2}dx$
$\frac{x^3-26x-41}{x+4}$
$\frac{dv}{dt}=125v$
$10xy+y-7yx-8y$
$-3x+2y=6$
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