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giúp mình điiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

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    Đáp án:

    `A=1+3+3^2+3^3+…+3^2023`
    `3A=3+3^2+3^3+3^4+…+3^2024`
    `3A-A=(3+3^2+3^3+3^4+…+3^2024)-(1+3+3^2+3^3+…+3^2023)`
    `2A=3^2024-1`
    `A=(3^2024-1)/2`
    Vậy: `A=(3^2024-1)/2`
    Thay `A` vào biểu thức `2A+1=81^x ,` ta có:
    `2*(3^2024-1)/2 +1=81^x`
    `3^2024-1+1=(3^4)^x`
    `3^2024=3^(4x)`
    `2024=4x`
    `2024:4=x`
    `506=x`
    Vậy: `x=506`

     

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