$34,521\cdot26,015$
$\lim_{x\to\infty}\left(1+4x\right)^{\frac{3}{x}}$
$2\ln\left(x-5\right)=\ln\left(x+13\right)+\ln13$
$18\le\frac{z}{3}+14$
$\left(4a-b-2c\right)^2$
$3x^2-5x=2$
$4a^2b-2ab+4ab-ab-3a^2b$
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