Rút gọn biểu thức : ( 3 + 1 ) ( 3^2 + 1 ) ( 3^4 + 1 ) ( 3^16 + 1 ) ( 3^32 + 1 )

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Rút gọn biểu thức :
( 3 + 1 ) ( 3^2 + 1 ) ( 3^4 + 1 ) ( 3^16 + 1 ) ( 3^32 + 1 )

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Sigridomena 1 year 2020-11-13T01:29:26+00:00 2 Answers 66 views 0

Answers ( )

    0
    2020-11-13T01:31:23+00:00

    Đáp án:

    $\dfrac{3^{64}-1}{2}$

    Giải thích các bước giải:

    Đặt: $A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$

    $⇔ 2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$

    $=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$

    $=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1)$

    $=(3^8-1)(3^8+1)(3^{16}+1)(3^{32}+1)$

    $=(3^{16}-1)(3^{16}+1)(3^{32}+1)$

    $=(3^{32}-1)(3^{32}+1)$

    $=3^{64}-1$

    $⇒ A=\dfrac{3^{64}-1}{2}$

    0
    2020-11-13T01:31:24+00:00

    Đáp án:

    $\dfrac{3^{64} -1}{2}$

    Giải thích các bước giải:

    $(3 +1)(3^2 + 1)(3^4 + 1)(3^8 +1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3 – 1)(3 +1)(3^2 + 1)(3^4 + 1)(3^8 +1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3^2 -1)(3^2 + 1)(3^4 + 1)(3^8+1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3^4- 1)(3^4 + 1)(3^8+1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3^8 – 1)(3^8+1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3^{16} – 1)(3^{16} + 1)(3^{32} + 1)$

    $= \dfrac{1}{2}(3^{32} -1)(3^{32} + 1)$

    $= \dfrac{3^{64} -1}{2}$

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