$\left(-\frac{3}{2}x^3+5\right)\left(\frac{1}{3}x^3-3\right)$
$\left(\frac{4}{5}yx^3+\frac{2}{3}y^4\right)^2$
$+4-15+6-20-3$
$7x\cdot8x$
$3m^2+3+m^2$
$\left(\sqrt{\frac{1}{k}}\right)$
$\frac{x}{6}-5=7$
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