$-\frac{dy}{dx}=3y\tan\left(x\right)=1$
$\left(2x+1\right)^2\le4x\left(x-3\right)+33$
$\frac{\left(x^2+4x+100\right)}{x+25}$
$\frac{\left(2a+3b\right)^{2}+\left(2a-3b\right)^{2}}{4a^{2}+9b^{2}}$
$2025:\left(+135\right)$
$\sqrt[3]{3^{27}}$
$-2g^2-24g=-26$
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