$3x^2-9<0$
$\left(-3x-4\right)\cdot\left(4x+5\right)$
$10a^4b^2-18a^2b^3$
$\left(a^2b^{11}+8\right)\left(a^2b^{11}-8\right)$
$\left|1+\left(-41\right)\right|\cdot\left(-9\right)$
$3.25+4.98$
$\frac{x}{x^3-x^2+x-1}$
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