$8x^2=2x+3$
$-15+\left(-9+10\right)^3-3\left(-8\right)^0\left(-2\right)^2$
$\ln\:\left(\left(\sec\:\left(x\right)+\tan\:\left(x\right)\right)^{\cos\:\left(x\right)}\right)$
$3x+8<2x+12$
$\tan\left(\frac{x}{2}\right)=\frac{\sec\left(x\right)-1}{\tan\left(x\right)}$
$\left(7x-3y\right)\left(4x-2y\right)$
$\frac{16a^2-49b^2}{4a-7b}$
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