$\left(6-4\ln\left(3-x\right)\right)^2$
$\frac{1}{4x^3+2x}dy-e^{-y}dx=0$
$x^2-4x+3=2$
$\frac{dy}{dx}=\frac{1+y^2}{y\sin\left(x\right)}$
$\frac{3x^2-19-6x+9x^4}{\left(3x^2-4\right)}$
$\left(-10\right)^6+\left(-10\right)^4$
$\frac{7x}{12}+\frac{5x}{12}$
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