Полное решение!11!Решите неравенство: [tex] \frac{ x^{2} -7x+6}{x-2} [/tex][tex] \leq 0[/tex]. Created by miass74rus. algebra-ru

Первое неравенство Получаем решение Второе неравенство Получаем решение Объединяя Новые вопросы в АлгебраДоведіть тотожність: (3а-4a² - ab - 4b²) + (4 - 3a a² - 2ab+7b2) = b(3а-4)(а-11b). дам 50 балів дуже срочноПреобразуйте в произведение 1-√2cosAнайдите остаток от деления суммы 7³ + 11 x 7 на 6Знайдіть шостий член геометричної прогресії, якщо b2=2, b5=16.Знайти суму членів геометричної прогресії від четвертого до восьмого включно, якщо b1=5, q=-2.Помогите решить 5 заданиеВ арифметичній прогресії a2+a5=11, різниця d=3. Визначте другий член a2 цієї прогресії.Бічна сторона рівнобедреного трикутника ділиться точкою дотику вписаного кола у відношенні 3 : 4, рахуючи вiд вершини кута при основі трикутника. Знай…діть бічну сторону трикутника, якщо його основа дорівнює 12 см.Бічна сторона рівнобедреного трикутника ділиться точкою дотику вписаного кола у відношенні 3 : 4, рахуючи вiд вершини кута при основі трикутника. Знай…діть бічну сторону трикутника, якщо його основа дорівнює 12 15/8+13/3. .........ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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\u043f\u0440\u043e\u0433\u0440\u0435\u0441\u0456\u0457."},{"id":49782113,"content":"\u0411\u0456\u0447\u043d\u0430 \u0441\u0442\u043e\u0440\u043e\u043d\u0430 \u0440\u0456\u0432\u043d\u043e\u0431\u0435\u0434\u0440\u0435\u043d\u043e\u0433\u043e \u0442\u0440\u0438\u043a\u0443\u0442\u043d\u0438\u043a\u0430 \u0434\u0456\u043b\u0438\u0442\u044c\u0441\u044f \u0442\u043e\u0447\u043a\u043e\u044e \u0434\u043e\u0442\u0438\u043a\u0443 \u0432\u043f\u0438\u0441\u0430\u043d\u043e\u0433\u043e \u043a\u043e\u043b\u0430 \u0443 \u0432\u0456\u0434\u043d\u043e\u0448\u0435\u043d\u043d\u0456 3 : 4, \u0440\u0430\u0445\u0443\u044e\u0447\u0438 \u0432i\u0434 \u0432\u0435\u0440\u0448\u0438\u043d\u0438 \u043a\u0443\u0442\u0430 \u043f\u0440\u0438 \u043e\u0441\u043d\u043e\u0432\u0456 \u0442\u0440\u0438\u043a\u0443\u0442\u043d\u0438\u043a\u0430. \u0417\u043d\u0430\u0439\u0434\u0456\u0442\u044c \u0431\u0456\u0447\u043d\u0443 \u0441\u0442\u043e\u0440\u043e\u043d\u0443 \u0442\u0440\u0438\u043a\u0443\u0442\u043d\u0438\u043a\u0430, \u044f\u043a\u0449\u043e \u0439\u043e\u0433\u043e \u043e\u0441\u043d\u043e\u0432\u0430 \u0434\u043e\u0440\u0456\u0432\u043d\u044e\u0454 12 \u0441\u043c.\u200b"},{"id":49782112,"content":"\u0411\u0456\u0447\u043d\u0430 \u0441\u0442\u043e\u0440\u043e\u043d\u0430 \u0440\u0456\u0432\u043d\u043e\u0431\u0435\u0434\u0440\u0435\u043d\u043e\u0433\u043e \u0442\u0440\u0438\u043a\u0443\u0442\u043d\u0438\u043a\u0430 \u0434\u0456\u043b\u0438\u0442\u044c\u0441\u044f \u0442\u043e\u0447\u043a\u043e\u044e \u0434\u043e\u0442\u0438\u043a\u0443 \u0432\u043f\u0438\u0441\u0430\u043d\u043e\u0433\u043e \u043a\u043e\u043b\u0430 \u0443 \u0432\u0456\u0434\u043d\u043e\u0448\u0435\u043d\u043d\u0456 3 : 4, \u0440\u0430\u0445\u0443\u044e\u0447\u0438 \u0432i\u0434 \u0432\u0435\u0440\u0448\u0438\u043d\u0438 \u043a\u0443\u0442\u0430 \u043f\u0440\u0438 \u043e\u0441\u043d\u043e\u0432\u0456 \u0442\u0440\u0438\u043a\u0443\u0442\u043d\u0438\u043a\u0430. 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tex]...

miass74rus @miass74rus

July 2022 1 6 Report

Полное решение!11!Решите неравенство: [tex] \frac{ x^{2} -7x+6}{x-2} [/tex][tex] \leq 0[/tex]

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Матов

Verified answer

$\frac{x^2-7x+6}{x-2} \leq 0\\\\
x^2-7x+6=(x-1)(x-6)\\\\\
 \frac{(x-1)(x-6)}{x-2} \leq 0\\\\
 x \neq 2\\\\

$
$1) \left \{ {{(x-1)(x-6) \geq 0} \atop {x-20}} \rigaht. \\\\$
Первое неравенство
$1) \left \{ {{(x-1)(x-6) \geq 0} \atop {x-2$
Получаем решение
$x \in (\infty;1)$

Второе неравенство
$\left \{ {{(x-1)(x-6) \leq 0 } \atop {x>2}} \right. \\\\
 \left \{ {{x \leq 1 ; \ \ x \geq 6} \atop {x>2}} \right. \\\\
 \left \{ {{x \geq 1; x \leq 6} \atop {x$
Получаем решение
$x \in (2;6]\\$

Объединяя
$x \in (-\infty;1) \ \ \cup \ \ \ (2;6]$

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