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$4x-\frac{3}{5}>-x-6$
$\sqrt[3]{48a^6b^9}$
$3x+2-5x+4-2x+1$
$7\:cos\left(x\right)+1=-3\:cos\:\left(x\right)+6$
$\left(7x+6\right)x\left(x^2+3\right)+2$
$\lim_{x\to\infty}\frac{\ln\left(3x\right)}{e^{7x}}$
$4x\:-\:9y\:=\:2\:$
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