$x-7=1$
$2x\left(3x+2\right)-8x<2x-2$
$4x-4+4x-4$
$\lim_{x\to4}\left(\frac{\left(6\sqrt{2}-2\right)}{3x-12}\right)$
$\frac{\left(x+1\right)}{\left(x-1\right)}+\frac{\left(2x\right)}{\left(x^2-1\right)}$
$x^2+37+70$
$4+2\cdot2$
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