$\:-\:\left(\:-\:\left(\:-10\:\right)\right)$
$3\left(4x-9\right)=x+6$
$\frac{2}{3}x+3>\frac{4}{3}x-1$
$x+3.4=9.1$
$7n-1=-8$
$6x^2-8x-8\:\cdot2x\:-4$
$\frac{1-tan^2\left(x\right)}{1+tan^2\left(x\right)}\:=\:\frac{2}{\cos^2\left(x\right)}-1$
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