Найдите количество целых решений неравенства: [tex]3 * 4^{x} - 5* 6^{x} + 2 * 9^{x} \leq 0[/tex]. Created by pomogite121415. algebra-ru

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December 2021 1 17 Report

Найдите количество целых решений неравенства:
[tex]3 * 4^{x} - 5* 6^{x} + 2 * 9^{x} \leq 0[/tex]

Answers & Comments

Alexandr130398 $3*4^x-5*6^x+2*9^x \leq 0 \\ \\ 3*4^x-5*6^x+2*9^x= 0 \ | :9^x, \ 9^x \neq 0 \\ \\ 3* (\frac{4}{9} )^x-5* (\frac{6}{9} )^x+2=0 \\ \\ 3* (\frac{2}{3} )^{2x}-5* (\frac{2}{3} )^x+2=0 \\ \\ (\frac{2}{3} )^x=t, \ t\ \textgreater \ 0 \\ \\ 3t^2-5t+2=0 \\ \\ D=25-24=1 \\ \\ t_1= \frac{5-1}{6}= \frac{2}{3} \\ \\ t_2= \frac{5+1}{6} =1 \\ \\ 1) (\frac{2}{3} )^x=\frac{2}{3} \\ \\ x=1 \\ \\ 2) \ (\frac{2}{3})^x=1 \\ \\ (\frac{2}{3})^x=(\frac{2}{3})^0 \\ \\ x=0$

$3*4^x-5*6^x+2*9^x \leq 0 \\ \\ +++++[0]----[1]+++++\ \textgreater \ x \\ \\ x \in [0;1]$

Целые решения: 0; 1
количество целых решений: 2

Ответ: 2

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