$-40x^2+-40x^2$
$-\left(x+3-x\right)=2x-7$
$-5tanx=6,5$
$\lim_{x\to\infty}\:\left(6+x^7\right)\left(\frac{1}{\ln\left(5x^2+1\right)}\right)$
$3.6^3$
$\frac{dy}{dx}=2xy-8x$
$\frac{dy}{dx}=\frac{2y-1}{5x-1}$
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