$ab-bc+ad-cd$
$\left(x+2\right)^2>\left(x+3\right)\left(x+2\right)$
$125^{2-x}\cdot5^x=1$
$\int x^2\left(4+3x^3\right)^5dx$
$4x\:+\:4\:+\:12\:-\:x$
$-3+\left(-3\right)^2$
$\frac{\tan^2\left(u\right)}{\sec\left(u\right)}=\sec\left(u\right)-\cos\left(u\right)$
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