$\frac{70\frac{\left(4x-3\right)-\left(8x+9\right)}{4x}}{\frac{4x}{4x}}$
$\lim_{x\to\infty}6x+3$
$1+\frac{\tan^2x}{1+\sec^2x}$
$18\le\frac{z}{3}+14$
$\frac{7}{1}$
$-4+2-8-3+3-5+4-10$
$2.282\left(4.7083\right)$
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