Simplifying
$\lim_{x\to\infty}\left(\frac{6x^2-3x+2}{4x-3}\right)$
$3x^2=4x$
$-2\left(-4uv\right)^2$
$\frac{x+1}{4}>-2$
$4x^2+2x-15=0$
$9\left(-2\cdot\left(-1\right)\cdot0\right)\cdot\left(-3\right)$
$\frac{2x\left(1-\left(4\right)+1\right)-2x}{x^{2}+y^{2}\sqrt{y^{2}+1}}$
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