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vaip4
@vaip4
July 2022
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Подскажите идею - как решать
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hammerfallwert23
2^x+3 + 2^x = 3^x2+2x-5 + 3^x2+2x-6
2^x(2^3+1)=3^x2+2x-6(3+1)
3^2 * 2^x = 2^2 * 3^x2+2x-6
(3^2)/(2^2)=(3^x2+2x-6)/(2^x)
x2+2x-6=2
x2+2x-8=0
x=-4 x=2
2^x=2^2
x=2
Ответ: x=2
2 votes
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vaip4
А ларчик то. Просто открывался) Благодраю
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Answers & Comments
2^x(2^3+1)=3^x2+2x-6(3+1)
3^2 * 2^x = 2^2 * 3^x2+2x-6
(3^2)/(2^2)=(3^x2+2x-6)/(2^x)
x2+2x-6=2
x2+2x-8=0
x=-4 x=2
2^x=2^2
x=2
Ответ: x=2