$\left(1-\frac{sin^2x}{16}\right)^{\frac{1}{2}}$
$6\le-\frac{5}{9}+x$
$4cos^2x-1=0$
$\frac{x^2-3x-10}{x}$
$\log_{11}\left(x^2+4x\right)=\log_{11}\left(5x+2\right)$
$\frac{4x^3+8x^2+7x+7}{2x+3}$
$246.24+238.78+98.3$
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