$\int\:\frac{\left(6x^2-3x+1\right)}{x^3-x^2}dx$
$\int_1^{\sqrt{3}}\left(\frac{1}{\left(x^2+1\right)}\right)dx$
$x^2\:+\:4y^2\:+\:5z^2\:=\:6\:$
$25x^220xy+16y^2$
$15x^2y-4x^2+6x^2-11x^2y$
$y'=\frac{6x-12}{x^2},y\left(1\right)=20$
$1500\le\frac{1500-450}{8}$
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