Question

помогите иррациональные неравенства
под корнем x + 12 < x

Answer 1

$sqrt{x+12} textless x; ; ; Leftrightarrow ; ; ; left{begin{array}{l}x+2 geq 0\x textgreater 0\x+12 textless x^2end{array}right ; ; left{begin{array}{l}x geq -2\x textgreater 0\x^2-x-12 textgreater 0end{array}right \\x^2-x-12 textgreater 0; ; ,; ; x_1=-3; ,; ; x_2=4\\+++(-3)---(4)+++quad xin (-infty ,-3)cup (4,+infty )\\ left { {{x textgreater 0} atop {xin (-infty ,-3)cup (4,+infty )}} right. ; ; Rightarrow ; ; ; ; underline {xin (4,+infty )}$