$4-x<2x-5$
$\int\left(\frac{u-8}{u^3-4u^2+4u}\right)du$
$\frac{5^{\frac{1}{2}}\cdot\:6^{\frac{1}{2}}}{12x^{\frac{1}{2}}}=\frac{\sqrt{5}\sqrt{6}\sqrt{x}}{12x}$
$0x\cdot1x$
$14-1-3$
$2\cdot\:\:cos\left(7x\right)+\sqrt{3}=0$
$-2\left(x+12\right)$
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