Помогите решить неравенство. Created by Vika878787. algebra-ru

Область определения неравенства: С учётом области определения: Ответ: malika1014malika1014Легче всего решить такОдну область потерялиХотя нет, вы правы)))))Новые вопросы в Алгебра1.Знайдіть область визначення функцiï f(x) =√(x-7)×(x+10)/x-22.Задайте функцію, обернену до функції у = 1/6x + 2.3.Побудуйте графік функції у =-1/2…√x +1. Користуючись побудованим графікомзнайдіть.1) нулi функцiï2) промiжки знакосталості:3) промiжки зростання і проміжки спадання4) область значень функцiï.допоможіть будь ласка Срочно1.42.13.1!!!!!!!!!!!допоможіть будь ласка 5.Знайдіть множину розв'язків нерівності:1)(x-6)(x+3)²(x+1)>02)x²+3x-10/x²+8x+16<06.Розв'яжіть нерівність:(x-4)²-(x+7)³/x-6 =>04.Дослідіть на парність функцію: 1)f(x)=4x⁷-2x³; 2) f(x)=x²-3x⁴: 3) f(x)=x³+x⁶решите уравнение срочно 3(х-5)+2(х+4)=3найдите допустимые значения переменной в вырожении1+3х²/(х²-16)(х+7)помогите, очень срочно(x-3) (x²+2x-9) =?8 клас (2x+3) (3y-1)=2x×3y-2x+3y-3=6xy-2x+÷×3y-3=6xy-2x+9y-3 ВІРНО??ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Помогите решить неравенство...

Vika878787 @Vika878787

July 2022 1 18 Report

Помогите решить неравенство

Answers & Comments

Apofeoz $\log_{\frac{1}{2}}(6-x)\ge\log_{\frac{1}{2}}x^2,$

Область определения неравенства: $1)\ 6-x>0,\\x0,\\x\in R.$

$\log_{2^{-1}}(6-x)\ge\log_{2^{-1}}x^2,\\-\log_{2}(6-x)\ge-\log_{2}x^2\
 |\bullet(-1),\\\log_2(6-x)\le\log_2x^2,\\\\6-x\le 
x^2,\\x^2+x-6\ge0,\\\\x^2+x-6=0\\D=1+24=25,\\\\x_1=\frac{-1+\sqrt{25}}{2}=2,\
 x_2=\frac{-1-\sqrt{25}}{2}=-3.\\\\(x+3)(x-2)\ge0,\\x\ge2,\ 
x\le-3.$

С учётом области определения: $\left \{ {{x\le-3,\ x\ge2} \atop {x\in R,\ x$

Ответ: $x\in(-\infty;\ -3]\wedge[2;\ 6).$