$y'=y\:\tanh x$
$\left(2x\:-\:4y\right)^2$
$7g+-3-2g+4^2$
$\frac{\frac{\left(2x-6\right)}{\left(y\right)}}{\frac{\left(x-3\right)}{y+1}}$
$\left(-2\right)\left(-2\right)\left(-2\right)\left(-2\right)\left(-2\right)-\:2$
$\left(4f+7\right)\left(4f-7\right)$
$7.50\cdot7.21$
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