$x^2-4b^2$
$z^24z+4$
$tan\left(x-\frac{3\pi}{4}\right)=\frac{tanx+1}{1-tanx}$
$\int_1^2\left(\frac{1}{\sqrt[2]{x^4-1}}\right)dx$
$\:\:y=\sqrt{\frac{4}{x-2}}$
$\left(-21\right)-\left(-10\right)$
$.5x+6=x-4$
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