ms hỏi lần đầu mog làm nhanh akkk

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ms hỏi lần đầu mog làm nhanh akkk
ms-hoi-lan-dau-mog-lam-nhanh-akkk

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Farah 4 years 2020-11-01T18:50:48+00:00 2 Answers 54 views 0

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  1. sori
    0
    2020-11-01T18:52:16+00:00
    Reply

    mình chỉ làm đc có z thôi :)))!

     

    ms-hoi-lan-dau-mog-lam-nhanh-akkk

  2. Jezebel
    0
    2020-11-01T18:52:30+00:00
    Reply

    Bài $1$.

    $a$) $4.5^3 – 17.2^4 + 35. 2020^0$

    $= 4.5^3 – 17.2^4 + 35.1$

    $= 4.125 – 17.16 + 35$

    $= 500 – 272 + 35$

    $= 263$.

    $b$) $10^3 – {5^2 . 10 : [75 – (7-2)^2]}$

    $= 1000 – [25.10 : (75-5^2)]$

    $= 1000 – [250 : (75-25)]$

    $= 1000 – (250 : 50)$

    $= 1000 – 5$

    $= 995$.

    Bài $2$.

    $a$) $390 – (x-7) = 169 : 13$

    $⇔ 390 – (x-7) = 13$

    $⇔ x – 7 = 377$

    $⇔ x = 384$

      Vậy $x=384$.

    $b$) $(x-140):7 = 3^3 – 2^3 . 3$

    $⇔ (x-140):7 = 27 – 8.3$

    $⇔ (x-140) : 7 = 3$

    $⇔ x – 140 = 21$

    $⇔ x = 161$

      Vậy $x=161$.

    $c$) $3^{x+5} – 2. 3^{x+2} = 225$

    $⇔ 3^{x+2} . 3^3 – 2 . 3^{x+2} = 225$

    $⇔ 3^{x+2} . (3^3 – 2) = 225$

    $⇔ 3^{x+2} . (27-2) = 225$

    $⇔ 3^{x+2} . 25 = 225$

    $⇔ 3^{x+2} = 9$

    $⇔ 3^{x+2} = 3^2$

    $⇔ x+2 = 2$

    $⇔ x = 0$

      Vậy $x=0$.

    Bài $3$

    $A = \dfrac{11 . 3^{22} . 3^7 – 9^{15}}{(2.3^{14})^2}$

    $⇔ A = \dfrac{11 . 3^{29} – (3^2)^{15}}{2^2 . 3^{28}}$

    $⇔ A = \dfrac{11.3^{29} – 3^{30}}{2^2 . 3^{28}}$

    $⇔ A = \dfrac{3^{29} . (11 – 3)}{2^2 . 3^{28}}$

    $⇔ A = \dfrac{3^{29} . 8}{2^2 . 3^{28}}$

    $⇔ A = \dfrac{3^{29} . 2^3}{2^2 . 3^{28}}$

    $⇔ A = \dfrac{3.2}{1} = 6$.

     

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