$\frac{4x^2-3}{x-5}$
$\frac{3x+1}{5}-\frac{1}{3}\le\frac{2}{15}\cdot\left(3x+2\right)+\frac{4\cdot\left(1-x\right)}{3}$
$2x^2+6y^2=1$
$-x^3+x^2=0$
$\left(x+2y+10z+3w\right)^2$
$3a\left(a^3-2a+5\right)$
$\lim_{x\to0}\left(\frac{\sin6}{\cos5}\right)$
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