$-18:-6$
$2x^2+8x=12$
$\sec^22x-1=\tan^22x$
$2\tan^2l+3\sec x=0$
$\int\left(\frac{x}{x^4+6}\right)dx$
$2\left(-14+r\right)-\left(-3r-0\right)$
$\frac{1}{8}x\frac{2}{3}$
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