$\frac{x-1}{6}=\frac{x-3}{4}$
$\frac{\left(2y^2-3\right)}{\left(y^2+1\right)}$
$-1+7-\left(-3\right)$
$\left(x-5\right)\left(x-5\right)$
$\lim_{x\to0}\frac{x^2+1}{x^2-1}$
$7.\left(4y\:+yx2\right)dy\:-\:\left(2x\:+xy2\right)dx\:=\:0$
$7 x - 4 = 3 x + 8$
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