Question

(b-5) в квадраті>b(b-10)

Answer 1

$\displaystyle\bf\\(b-5)^{2} > b(b-10)\\\\b^{2} -10b+25 > b^{2} -10b\\\\b^{2} -10b-b^{2} +10b > -25\\\\0\cdot b > -25\\\\b\in(-\infty \ ; \ +\infty)$