Помогите пожалуйста!!!. Created by yuliyakhayrulina. algebra-ru

Помогите пожалуйста!!!...

July 2022 1 1 Report

Помогите пожалуйста!!!

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ukinbid Log(x-3) (x-6)^2<=0
9^x-2-37*3^x-3+30<=0
1)log(x-3) (x-6)^2<=0
ОДЗ
x-3≠1;x≠4
x-3>0;x>3
(x-6)^2>0;x≠6
x∈(3;4)(4;6)(6;+∞)
Если 0<x-3<1;
3<x<4
x^2-12x+36>=1
x^2-12x+35>=0
(x-7)(x-5)>=0
x=5;7
\\\\\\\\\\\(+)\\\\\\\\\(5)------(-)----(7)\\\\\\(+)\\\\\\\\>
----(3)\\\\\\\\\(4)---------------------------------------->
x∈(3;4)
Если x-3>=1
x>=4
--------(+)--------(5)\\\\\\\(-)\\\\\\\\\(7)------(+)---->
-----------(4)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\>
x∈[5;7]
2)9^x-2-37*3^x-3+30<=0
3^2(x-2)-37*3^-1*3^x-2+30<=0
y^2-37/3y+30<=0
D=1369/9-120=289/9
y1;2=(37/3+-17/3)/2=10/3;9
3^x-2=9
x-2=2
x=4
3^x-2=10/3
log3 3^x-2=log3 10/3
x-2=log3 10/3
x=log3 10+1=log3 30
log3 27<log3 30<log3 81
3<log3 30<4
-------------(log3 30)\\\\(-)\\\(4)------->
x∈[log3 30; 4]
3)-------------(log3 30)\\\\(-)\\\\\\(4)------->
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(4)------(5)\\\\\\\\\\\\(7)-->
----------(3)\\\\\\\\\\\\\\\\\\\\\\\\\\\\(4)---------->
x∈[log3 30; 4)
Ответ [log3 30; 4) или [1+log3 10; 4)

2 votes Thanks 8

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