помогите, очень срочно!!!!. Created by Dfgh11. algebra-ru

помогите, очень срочно!!!!...

Dfgh11 @Dfgh11

November 2021 1 3 Report

помогите, очень срочно!!!!

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sedinalana

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1
ОДЗ
(x-3)/(x-4)>0
x=3 x=4
x<3 U x>4
x∈(-∞;3) U (4;∞)
log₇[(x-3)(x-4)]≤log₇[49(x-3)/(x-4)]
(x-3)(x-4)≤49(x-3)/(x-4)
(x-3)(x-4)-49(x-3)/(x-4)≤0
[(x-3)(x-4)²-49]/(x-4)≤0
(x-3)(x²-8x+16-49)/(x-4)≤0
(x-3)(x²-8x-33)/(x-4)≤0
(x-3)(x-11)(x+3)/(x-4)≤0
x=3 x=11 x=-3 x=4
+ _ + _ +
--------------[-3]------------[3]----------------(4)---------------[11]------------------
x∈[-3;3] U (4;11]
2
ОДЗ x>y
{2+log₂(x-y)=3⇒log₂(x-y)=1⇒x-y=2⇒x=2+y
{log₂(x²-y²)=3⇒x²-y²=8
(2+y)²-y²=8
4+4y+y²-y²=8
4y=4
y=1
x=3
3
x>0,x≠1
x^log₃x=x⁴/27
27^x^log₂x=x⁴
Прологарифмируем по основанию 2
27log²₂x-4log₂x=0
log₂x(27log₂x-4)=0
log₂x=0⇒x=1не удов усл
log₂x=4/27⇒ $x= 2^{4/27}$

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privedite dva primera dokazyvayushie pravilnost vyskazyvaniya vybirat sebe prav

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Dfgh11 July 2022 | 0 Ответы

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privedite dva primera dokazyvayushie pravilnost vyskazyvaniya vybirat sebe prav

kakuyu problemu podnimaet volter v dannom vyskazyvanii gosudarstvo delaet slaby

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