Алгебра. Условие на фото. Created by nastenkakrupkoq. algebra-ru

Алгебра. Условие на фото...

nastenkakrupkoq @nastenkakrupkoq

July 2022 1 1 Report

Алгебра. Условие на фото

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$\frac{(x+6)^2(x-1)^2}{(x^2+5x-24)^2}\geq \frac{2x^2+10x-12-x^2-5x+24}{x^2+5x-24}\\\frac{(x^2+5x-6)^2}{(x^2+5x-24)^2}\geq \frac{x^2+5x+12}{x^2+5x-24}$

Пусть $t=x^2+5x-24$

$\frac{(t+18)^2}{t^2}\geq \frac{t+36}{t}\\\frac{t^2+36t+324}{t^2}-\frac{t^2+36t}{t^2}\geq 0\\\frac{324}{t^2}\geq 0 \Leftrightarrow t\neq 0\\(x-3)(x+8)\neq 0\\x\neq -8;\ 3$

$|x|\leq 9 \Leftrightarrow -9\leq x\leq 9$ .

Сумма целых решений от -9 до 9 равна:

-9-7-6-5-4-3-2-1+0+1+2+4+5+6+7+8+9 = -3+8 = 5

Ответ: 5

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