$f\left(x\right)=\left(x-4\right)\left(3x+1\right)$
$\lim\:_{x\to\:\infty\:}\left(\sqrt{x}\cdot\:\frac{\left(4x+1\right)}{\sqrt{x^3+6x^2+9x}}\right)$
$1m-2+3m-6$
$-25x=-25$
$-y^2dv+y\sqrt{v}dy$
$y=x^7\cdot x^6$
$x^2+26x+14$
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