$\frac{\left(4x^3-3x^2\right)}{x^2}$
$x^3-7x^2+5x+25$
$+2+\left(+8\right).\left(6\right)-\left(-7+6\right)$
$9-5x+9y=0$
$4\frac{x^2}{y}+2\frac{x^2}{y}-5\frac{x^2}{y}$
$-\frac{13}{10}\left(a+\frac{10}{39}b\right)$
$\tan\left(x\right)\cot\left(x\right)=\cos\left(x\right)$
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