$24+0,85+3,1+5,128$
$\left(n^6\right)^8$
$\frac{dq}{dp}=\frac{2p}{2q+1}$
$4\cdot\left(x+5\right)=\left(3x+1\right)\cdot3+2$
$\log\left(x^2-4\right)-\log\left(x+2\right)=\log2$
$24-3y+y+7$
$\left(2sen\left(x\right)-\sqrt{3}\right)\left(2cos\left(x\right)-\sqrt{2}\right)=0$
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