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)