Найти область определения функции:. Created by ArtakN. algebra-ru

1x²+5x+6≥0x1+x2=-5 U x1*x2=6⇒x1=-3 U x2=-2x∈(-∞;-3] U [-2;∞)2(9-x²)/(x-2)>0(3-x)(3+x)/(x-2)>0x=3 x=-3 x=2 + _ + _-------------(-3)---------------(2)----------(3)-----------------x∈(-∞;-3) U (2;3)3{36-x²≥0⇒(6-x)(6+x)≥0 x=6 U x=-6 -6≤x≤6{x+5>0⇒x>-5{log(22)(x+5)≠0⇒x+5≠1⇒x≠-4x∈(-5;-4) U (-4;6]4{-x²+x+2≥0⇒x²-x-2≤0 x1+x2=1 U x1*x2=-2 -1≤x≤2{cosx>0⇒-π/2+2πn<x<π/2+2πn,n∈zx∈(-π/3;π/2)Новые вопросы в АлгебраПомогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х=-1 у=5 то 2х+9у-7Помогите решить. х÷х+9=14ПЛС М+ПОМОГИТЕ, ДАМС КОРОНУ И МНОГО 30 БАЛЛОВ(Нужно объяснение)557,559 номера. Если возможно,с решением Разложить на множетелиy^8-y^4+4y^2-4Дайте відповідь з рішенням будь ласка(557,559)1) 3 1/9 : 2 1/3 - 2 5/6 =28/9 * 3/7 - 17/6=4/3-17/6= 8/6-17/6=-3/2=-1.5как он получил -3/2Іван зірвав на клумбі 8 нарцисів та 4 тюльпани. Скільки всього існуєспособів вибору із цих квітів 3 нарцисів та 2 тюльпанів для букета?Укажіть функцію, графіком якої є парабола з вершиною у точці О(- 2; 0)ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Найти область определения функции:...

ArtakN @ArtakN

August 2022 1 11 Report

Найти область определения функции:

Answers & Comments

sedinalana

Verified answer

1
x²+5x+6≥0
x1+x2=-5 U x1*x2=6⇒x1=-3 U x2=-2
x∈(-∞;-3] U [-2;∞)
2
(9-x²)/(x-2)>0
(3-x)(3+x)/(x-2)>0
x=3 x=-3 x=2
+ _ + _
-------------(-3)---------------(2)----------(3)-----------------
x∈(-∞;-3) U (2;3)
3
{36-x²≥0⇒(6-x)(6+x)≥0 x=6 U x=-6 -6≤x≤6
{x+5>0⇒x>-5
{log(22)(x+5)≠0⇒x+5≠1⇒x≠-4
x∈(-5;-4) U (-4;6]
4
{-x²+x+2≥0⇒x²-x-2≤0 x1+x2=1 U x1*x2=-2 -1≤x≤2
{cosx>0⇒-π/2+2πn<x<π/2+2πn,n∈z
x∈(-π/3;π/2)