$\lim_{x\to\infty}\left(\ln\left(x\right)\right)^{\frac{1}{1-\ln\left(x\right)}}$
$6x^2+8x=8$
$0.5x^2\left(x^3-8x^2-1\right)$
$36x^6y^4-0.25$
$12-\left(-3\cdot4\right)-\left(5\cdot2\right)$
$2x^3+3y^2=5$
$-25\:+\:\left(-6\right)\:-\:\left(-1\right)\:$
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