$\lim_{x\to\infty}\frac{1-x}{\sqrt{1-z^2}}$
$\sin\left(bx\right)\sin\left(3x\right)$
$\sqrt{3x=12}$
$y=\frac{\left(2x-1\right)^2\left(x^2+3\right)^3}{\left(x+1\right)^2}$
$\frac{d}{dx}4x^3+3y^3=7$
$a+b+c=2$
$\frac{dy}{dx}+y\cos\left(x\right)=4\cos\left(x\right)$
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