Задача 57660 x^2+400/(x-7)^2 меньше или равно 14x-8...
Условие
Решение
[m]x^2-14x+49+ (\frac{20}{x-7})^2 ≤ -8+49[/m]
[m](x-7)^2+ (\frac{20}{x-7})^2 ≤ 41[/m]
Применяем формулу:
[r]a^2-2ab+b^2=(a-b)^2[/r]
[m](x-7)^2-40+ (\frac{20}{x-7})^2 ≤ 41-40[/m]
[m](x-7- \frac{20}{x-7})^2 ≤ 41-40[/m]
[m](x-7- \frac{20}{x-7})^2 ≤1[/m]
[m](x-7- \frac{20}{x-7})^2-1 ≤0[/m]
[m](x-7- \frac{20}{x-7}-1 )(x-7- \frac{20}{x-7}+1)≤0[/m]
[m]\frac{(x^2-15x+36)(x^2-13x+22)}{(x-7)^2}≤0[/m]
D_(1) =(-15)^2-4*36 =225-144=81>0
D_(2) =(-13)^2-4*22=169-88=81 >0
[m]\frac{(x-3)(x-12)(x-2)(x-11)}{(x-7)^2}≤0[/m]
Решаем методом интервалов:
______ [2] _____ [3] __________ (7) _______ [11] ______ {12] ___[red]+[/red]__
___+___ [2] ___[green]-[/green]__ [3] ____+______ (7) ____+___ [11] ___[green]-[/green]___ {12] ___[red]+[/red]__
О т в е т. [2;3] U [11;12]