Ребят, прошу помощи!!!!Одна система!У меня ЧП!!!!! Помогите!!!!!сижу уже больше 2 часов!!!!!!. Created by abih98. algebra-ru

1)2)ОДЗ: x+1>0 x> -12y²-2x+4=4(x+1)2y²-2x-4x=4-42y²-6x=0y²-3x=0y²=3x{y²=x+2{y²=3xx+2=3xx-3x=-2-2x=-2x=1y²=3*1y²=3y₁=√3y₂= -√3Ответ: (1; -√3); (1; √3).maryfinemaryfineРешение в прикрепленном файлеНовые вопросы в АлгебраСтепень с целым показателем. Свойства степени с целым показателем. Упростите.помогите пжпжппжпж1.7-0.3m=2-1.7mможете решить1.7-0.3m=2-1.7решите пожалуйстаСколькими способами можно рассадить 5 мальчиков и 3 девочки на 8 стульях, стоящих в один ряд, чтобы все девочки не сидели рядом. Пожалуйста, помогите…решить.3×(a-b-c)+5×(a-b)+3c=8×(a-b)Допоможіть будь ласка!Якщо можна з поступовим поясненнямДякую!m+3 n+3 якщо m>n порівняйте числазависимость y от х задана таблицей. Найдите ее область определения и область значения. Определите, является ли эта зависимость функцией. Постройте гра…фик этой зависимостих -4 -1 3 4 -4у -3 -2 1 -2 3найти производную соs²xНадо решить системы уравнений ( не графичным способом)125a⁸b⁴-20a⁶b⁵______________15a⁵b⁴ПредыдущийСледующий Вместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Ребят, прошу помощи!!!!Одна система!У меня ЧП!!!!! Помогите!!!!!сижу уже больше ...

abih98 @abih98

August 2022 1 19 Report

Ребят, прошу помощи!!!!Одна система!У меня ЧП!!!!! Помогите!!!!!сижу уже больше 2 часов!!!!!!

Answers & Comments

maryfine 1)
$( \frac{1}{36} )^{-y^2}=6^{2x+4} \\ \\ 
(6^{-2})^{-y^2}=6^{2x+4} \\ \\ 
6^{2y^2}=6^{2x+4} \\ \\ 
2y^2=2x+4 \\ 
2y^2=2(x+2) \\ 
y^2=x+2$

2)
ОДЗ: x+1>0
x> -1
$log_{4}(2y^2-2x+4)=6^{log_{6}(3lg \sqrt[3]{10} )}+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=3lg \sqrt[3]{10}+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=lg( \sqrt[3]{10} )^3+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=lg10+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=1+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=log_{4}4+log_{4}(x+1) \\ \\ 
log_{4}(2y^2-2x+4)=log_{4}(4(x+1))$
2y²-2x+4=4(x+1)
2y²-2x-4x=4-4
2y²-6x=0
y²-3x=0
y²=3x

{y²=x+2
{y²=3x

x+2=3x
x-3x=-2
-2x=-2
x=1

y²=3*1
y²=3
y₁=√3
y₂= -√3

Ответ: (1; -√3);
(1; √3).