$x^2+2x+4=7$
$cos\frac{1}{4}x=\frac{-\sqrt{2}}{2}$
$\frac{dy}{dx}xy=6$
$\frac{du}{dx}-3u=3x$
$\int\left(1+4x\right)^{\frac{-1}{2}}dx$
$5-x\ge-2$
$3x^2-8x+16$
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