$\left(3x-\frac{1}{4}\right)^2$
$-5\left(x-10\right)^2$
$\int\left(\frac{1}{\left(1+6t\right)}\right)dt$
$\left(-3469\right)\cdot7$
$2\cos^2x-5\cos x+2=0$
$\frac{\frac{1}{4}x^2y^3}{\frac{2}{3}x^5y^2}$
$16y^2-81$
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