помогите решить плиз. Created by olechka199812. algebra-ru

ОДЗ: x+1≥0 x≥ -1ОДЗ: y≠0Разложим на множители:y²-y-12=0D=1+48=49y₁=(1-7)/2=-3y₂=(1+7)/2=4y²-y-12=(y+3)(y-4)y(y+3)(y-4)<0y=0 y=-3 y=4 - + - + ------- -3 ---------- 0 ----------- 4 -----------\\\\\\\\\ \\\\\\\\\\\\\y=5 + + + | +y=1 + + - | -y= -1 - + - | +y= -4 - - - | -{y< -3{ 0<y<41) y < -3нет решений.2) 0<y<4{x≥ -1{x<3 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\-------- -1 ------------ 3 ----------\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x∈[-1; 3)Ответ: [-1; 3).y=5 + + + | +y=1 + + - | -y= -1 - + - | +y= -4 - - - | - что этоэто определение знака "+" или "-" на интервалах спасибопомогите пожалуйста http://znanija.com/task/17128010Новые 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______________15a⁵b⁴ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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olechka199812 @olechka199812

August 2022 1 21 Report

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m11m

Verified answer

ОДЗ: x+1≥0
x≥ -1

$2^{ \sqrt{x+1} }-1\ \textless \ \frac{3*2^2}{2^{\sqrt{x+1}}} \\ \\ 
y=2^{\sqrt{x+1}} \\ \\ 
y-1\ \textless \ \frac{12}{y} \\ \\ 
y-1- \frac{12}{y}\ \textless \ 0$

ОДЗ: y≠0
$\frac{y^2-y-12}{y}\ \textless \ 0$
Разложим на множители:
y²-y-12=0
D=1+48=49
y₁=(1-7)/2=-3
y₂=(1+7)/2=4

y²-y-12=(y+3)(y-4)

y(y+3)(y-4)<0
y=0 y=-3 y=4
- + - +
------- -3 ---------- 0 ----------- 4 -----------
\\\\\\\\\ \\\\\\\\\\\\\
y=5 + + + | +
y=1 + + - | -
y= -1 - + - | +
y= -4 - - - | -

{y< -3
{ 0<y<4

1) y < -3
$2^{ \sqrt{x+1} }\ \textless \ -3$
нет решений.

2) 0<y<4
$0\ \textless \ 2^{ \sqrt{x+1} }\ \textless \ 4 \\ 

a) 2^{ \sqrt{x+1} }\ \textgreater \ 0 \\ 
 x \geq -1 \\ \\ 
b)2^{ \sqrt{x+1} }\ \textless \ 4 \\ 
2^{ \sqrt{x+1} }\ \textless \ 2^2 \\ 
 \sqrt{x+1}\ \textless \ 2 \\ 
x+1\ \textless \ 2^2 \\ 
x+1\ \textless \ 4 \\ 
x\ \textless \ 4-1 \\ 
x\ \textless \ 3$

{x≥ -1
{x<3
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------- -1 ------------ 3 ----------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[-1; 3)
Ответ: [-1; 3).

y=5 + + + | +
y=1 + + - | -
y= -1 - + - | +
y= -4 - - - | - что это
это определение знака "+" или "-" на интервалах
спасибо
помогите пожалуйста http://znanija.com/task/17128010

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