$3x+9>5$
$\sqrt{9a^6b^8}$
$\frac{6}{a-8}=\frac{-a}{a-4}$
$\left(-3\right)\left(x+4\right)$
$cos\left(x\right)^2=\frac{1-sin\left(x\right)^2}{cos\left(x\right)}$
$-5x\le20$
$3a\frac{2}{3}a\frac{4}{3}$
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