Помогите решить 80 баллов а)7*3^x+1-3^x+4=5^x+2-5^x+3 б)2^x-2^x+2+2^x-1=(3^x+1-3^x+2+3^x)*2/9. Created by Andylov. algebra-ru

9...

Andylov @Andylov

August 2022 1 28 Report

Помогите решить 80 баллов
а)7*3^x+1-3^x+4=5^x+2-5^x+3
б)2^x-2^x+2+2^x-1=(3^x+1-3^x+2+3^x)*2/9

Answers & Comments

90misha90

Verified answer

$7*3^{x+1}-3^{x+4}=5^{x+2}-5^{x+3}$
$7*3*3^{x}-3^4*3^{x}=25*5^{x}-125*5^{x}$
$-60*3^{x}=-100*5^{x}$
$-60*3^{x}=-100*5^{x}$
$3*3^{x}=5*5^{x}$
$3^{x+1}=5^{x+1}$
$( \frac{3}{5}) ^{x+1}=1$
$( \frac{3}{5}) ^{x+1}= (\frac{3}{5} )^0$
$x+1=0$

$x=-1$
----------------------------------------------------------------------------
$2^x-2^{x+2}+2^{x-1}=(3^{x+1}-3^{x+2}+3^x)* \frac{2}{9}$
$2^x-4*2^{x}+ \frac{1}{2}* 2^{x}=(3*3^{x}-9*3^{x}+3^x)* \frac{2}{9} |*2$
$2*2^x-8*2^{x}+ 2^{x}=(3*3^{x}-9*3^{x}+3^x)* \frac{4}{9}$
$-5*2^{x}=(-5*3^x)* \frac{4}{9}$
$2^{x}=3^x *(\frac{2}{3})^2$
$(\frac{2}{3})^x =(\frac{2}{3})^2$

$x=2$

2 votes Thanks 1

Answers & Comments

Verified answer

pomogite reshitdayu 40 ballove70b5f3c86e120e159133be980a87efe 37990

najdite oblast opredeleniya funkcii vnutri zadanie6 33 bza pravilnyj otvet

pomogite s resheniemdayu 40 ballov

5 sdelat 25

nuzhno sostavit krossvord na temu okonchanie i osnovapomogite s voprosami

pomogite reshitfoto vnutrinuzhno sdelatkotorye ne vydelennye kruzhochkom

pomogitedayu 50 ballovnajdite izobrazhenie predmeta a v sobirayushej linze te

iz kakih stran vyezzhayut bolshe vsego turistov10 stran dayu 100 ballov

nuzhna pomosh v reshenie primerabudu priznatelna za resheniepravilnyj otvet sd

postrojte izobrazhenie predmeta v rassveivayushej linzepomogitedayu 40 ballov

рекомендуемые вопросы

Helpful Links

Helpful Social