$y^2+2x=2$
$y^{\frac{1}{4}}=xk$
$\frac{2z^2}{3y}.\frac{5z}{7}$
$\frac{1}{\left(x^2\left(\sqrt{x^2-4}\right)\right)}$
$\int_0^2kx^2\left(2-x\right)dx$
$1x^2-0,6x-0,2$
$-9x^2+x+5x^5$
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