$3b^2+2b^3-2b^2$
$g\left(x\right)=\left(x^2+1\right)^3\left(x-1\right)^7x^3$
$3-7x\geq 4$
$\frac{6x^4-12x^3+15x^2+5x-3}{3x^2-1}$
$\lim_{x\to\infty}\frac{9x^5+3x^2}{4-x^2}$
$2x^2+13x-24=0$
$\lim_{x\to4}\sqrt{x+4}$
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