Question

I. Степенная функция. 10 класс.(задание на фото)

Answer 1

$sqrt{x+1}<x-1$ ⇔ $left { {{x+1<(x-1)^2} atop {x+1 geq 0}} atop {x-1>0}} right.$ ⇔ $left { {{x+1<x^2-2x+1} atop {x geq-1}} atop {x>1}} right.$ ⇔ $left { {{x^2-2x+1-x-1>0} atop {x geq-1}} atop {x>1}} right.$ ⇔ $left { {{x^2-3x>0} atop {x geq-1}} atop {x>1}} right.$ ⇔ $left { {{x(x-3)>0} atop {x geq-1}} atop {x>1}} right.$
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Ответ: x∈(3;+бесконечность)