Очень срочно, помогите пожалуйста, очень нужно. Created by annagrishchenko. algebra-ru

Log(2)(x²/4)=log(2)x²-log(2)4=2log(2)x-2log(0,5)(0,5x)=log(0,5)0,5+log(0,5)x=1+log(0,5)x=1-log(2)xlog(3)(x/2)/log(3)2=log(2)(x/2)=log(2)x-log(2)2=log(2)x-1получаем2(log(2)x-1)*(1-log(2)x)≤log(2)x-1-2(log(2)x-1)²-(log(2)x-1)≤0(log(2)x-1)³≥0{x>0{log(2)x-1≥0⇒log(2)x≥1⇒x≥2x∈[2;∞)Новые вопросы в АлгебраСрочно1.42.13.1!!!!!!!!!!! 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??Помогите с системойпожалуйстаh=0.2мP=1000кг/м^3P=900 кг/м^3h -?ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Очень срочно, помогите пожалуйста, очень нужно...

annagrishchenko @annagrishchenko

July 2022 1 39 Report

Очень срочно, помогите пожалуйста, очень нужно

Answers & Comments

sedinalana

Verified answer

Log(2)(x²/4)=log(2)x²-log(2)4=2log(2)x-2
log(0,5)(0,5x)=log(0,5)0,5+log(0,5)x=1+log(0,5)x=1-log(2)x
log(3)(x/2)/log(3)2=log(2)(x/2)=log(2)x-log(2)2=log(2)x-1
получаем
2(log(2)x-1)*(1-log(2)x)≤log(2)x-1
-2(log(2)x-1)²-(log(2)x-1)≤0
(log(2)x-1)³≥0
{x>0
{log(2)x-1≥0⇒log(2)x≥1⇒x≥2
x∈[2;∞)