$+4-1+6$
$\left(1-\sin^2x\right).\csc^2x=1$
$4y=x^2+10x+41\:$
$\:9x-3+3x=129$
$\left(-\frac{1}{2}n^2\right)\cdot\left(+\frac{3}{2}m^2\right)$
$\log\left(5x+1\right)+\log3$
$-6-4\left(x-3\right)+6x>2+3x$
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