$\left(\frac{2x^{-4}y^3\cdot y}{\left(2x^0y^{-4}\right)^{-1}}\right)$
$2x^2-3x+4\:-1x^2+6x$
$-15\:-9\:+\:2\:+\:-\:13$
$\left(2x+4\right)\left(x-1\right)$
$\int_1^{\infty}\ln\left(x^2\right)dx$
$-3+4\sqrt{3}tgx=-15$
$2a\left(3x+y\right)+8\left(3x+y\right)-10a\left(3x+y\right)$
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