√4x+1>7 Пж помогите решить , подробно
7\\\\4x+1>7^{2}\\\\4x+1>49\\\\4x>48\\\\x>12" alt="\sqrt{4x+1}>7\\\\4x+1>7^{2}\\\\4x+1>49\\\\4x>48\\\\x>12" align="absmiddle" class="latex-formula">
Ответ : x ∈ (12 , + ∞)
7\\\\4x+1>7^{2}\\\\4x+1>49\\\\4x>48\\\\x>12" alt="\sqrt{4x+1}>7\\\\4x+1>7^{2}\\\\4x+1>49\\\\4x>48\\\\x>12" align="absmiddle" class="latex-formula">