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$\frac{x^2+x}{4x}$
$\left(\frac{1}{4}-x\right)\left(\frac{3}{2}-x\right)$
$2x-6>4$
$\frac{26-x}{x}=12$
$20-4x=2x-10$
$\log\left(128-7x\right)=2$
$\lim_{x\to\infty}\left(1+\frac{12}{x}\right)^{\frac{x}{8}}$
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