$\left(-3x^2+4x-7\right)-\left(x^3+4x^2-\frac{3}{4}x\right)$
$12x^2+16x+4$
$\left(4z^2+6a^2\right)^5$
$y^2+3y=1$
$6-5\left(1-x\right)\le4$
$y'-3x^2y=4xe^{x^3}$
$-3x+\frac{1}{2}\le4$
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