$10x\:-\:12y\:+\:8\:+12y$
$\sin x^2+3\sin x\cos x+\cos^2x=0$
$-9\cdot6+3+8$
$\int\left(2x-7\right)^{-5}dx$
$x^3-4x^2-6x+5$
$\left(\frac{3}{5}x-\frac{4}{7}\right)^2$
$\int\left(\frac{6}{\left(t+9\right)^9}\right)dt$
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