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yurmalnikroman
@yurmalnikroman
July 2022
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пожалуйста Решите уравнение:↓☻
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QuasarDreemov
Запишем данное уравнение в таком виде (log2(x))^2-log0,5(1/x)=6; {х>0, х не=0; <=> (log2(x))^2-log2^-1(x^-1)=6 (0,5=2^-1=1/2=0,5; 1/x=x^-1) <=> (log2(x))^2-log2(x)-6=0. Пусть log2(x)=t,
Имеем: t^2-t-6=0; D=(-1)^2+24=25; [t1=3, t2=-2. Возвращаемся к замене: а) log2(x)=3; x=2^3=8; б) log2(x)=-2; x=2^-2=1/4.
1) [(log2(8))^2-log2(8)=6] <=> [3^2-3=6] <=> [6=6];
2) [(log2(2^-2))^2-log2(2^-2>=6] <=> [(-2)^2+2=6] <=> [6=6].
Ответ: х=8, х=1/4.
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Имеем: t^2-t-6=0; D=(-1)^2+24=25; [t1=3, t2=-2. Возвращаемся к замене: а) log2(x)=3; x=2^3=8; б) log2(x)=-2; x=2^-2=1/4.
1) [(log2(8))^2-log2(8)=6] <=> [3^2-3=6] <=> [6=6];
2) [(log2(2^-2))^2-log2(2^-2>=6] <=> [(-2)^2+2=6] <=> [6=6].
Ответ: х=8, х=1/4.