$\left(x^2+3\right)\left(x^2+3y\right)$
$441^2+42x+1$
$\lim_{x\to3}\frac{x^2-11x+24}{x^2-6x+9}$
$f\left(x\right)=\sqrt[5]{\left(2x^3-x\right)^4}$
$10^2+\left(6-4\right)\left(5-2\right)^2$
$-36:\:6\:+\:4\cdot\left(-3\right)$
$f^2+7f+16$
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