$7\tan\theta-8=5\tan\theta-5$
$\frac{\left(2x^3\right)\left(12x^5\right)}{4x^2}$
$xy'+3y=x^6$
$\int_4^0\left(\frac{3x-6}{2}\right)dx$
$9\left(13-x\right)-4x=5\left(21-2x\right)+9x$
$-6\left(2r-2\right)=-8r+40$
$+1203-2$
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