$\frac{x}{4}=7$
$\left(4a^2+5xy^3\right)^2$
$3x\:-8\:+9\:-5x$
$\left(\frac{ac^4}{3b^3}\right)^2$
$y'+\left(x+3y\right)y$
$f\left(u\right)=\left(u-3\right)^2+3$
$log\left(4x\right)=log\left(x-3\right)$
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