$6x^2-5\left(x-1\right)=x\left(x+1\right)+4$
$\frac{-64}{-1}$
$9-3+4\cdot6$
$\lim_{x\to\infty}\left(\frac{\left(5x+6\right)}{8x^2+7x-7}\right)$
$2c^2+7c+3=0$
$d y - ( y - 8 ) ^ { 2 } d x = 0$
$16k^2+96k^3+144k^2-72k^2-216k+81$
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