$10xy\:y'=1-y^2$
$49a^4-70a^2+25$
$10+\left(4-21+12\right)-\left(-1-2\right)=\:\:$
$\int\frac{1}{\frac{9}{25}-x^2}dx$
$\left(x+1\right)\left(x+3\right)=0$
$1-\cos\left(x\right)^2+\cos\left(x\right)^4$
$2\cdot2\cdot3x^3x^5$
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