$2x^2-5x+6-5x^2+2x+4$
$\:\left(y-\frac{1}{3}\right)^2$
$\sqrt[3]{\left(2^2x^5\right)}\cdot\sqrt{\left(4^5x^3\right)}$
$7x+9\ge-x$
$y'\:+\:tan\left(x\right)y\:=\:2\:sin\left(x\right)\:cos^2\left(x\right)$
$\left(3x-2\right)\cdot\left(x^2-2x+3\right)$
$9^8\cdot9^{-5}$
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