$15\cdot\left(-25\right)$

$\left(+51\right).\left(-13\right).\left(-9\right)$

$\lim_{x\to\infty}\left(\frac{\tan\left(\frac{1}{x}\right)^2}{\ln\left(1+\frac{4}{x}\right)^2}\right)$

$9h^2+7,3h-12,9h^2+7,4-3,1h^3-5,2$

$x^2-1>\frac{3}{4}$

$6\left(2x-7y\right)+2y$

$\left(y^{m+2}+y^m\right)^2$