$\frac{7r^2}{4r^3}$
$40.375+26.5625+0.125+0.514285$
$\lim_{x\to0}\left(\frac{5\left(e^x-1\right)}{x^2-5x}\right)$
$\left(2d^3b\right)^4\cdot b^3$
$\int\frac{x^2-6x+9}{x^3+x^2-6x}dx$
$3x+1\le7x-2$
$\left(2a-4\right)\left(2a+3\right)$
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