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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: a,4^10.8^7/2^40 b 9^10.27^7/3^40 c 8^8.49^6/14^14.2^10

Tháng Hai 19, 2022 Toán học 2 Comments 0 views

Toán Lớp 7: a,4^10.8^7/2^40 b 9^10.27^7/3^40 c 8^8.49^6/14^14.2^10

Comments ( 2 )

  1. haianhtu97
    haianhtu97
    19 Tháng Hai, 2022 at 11:05 sáng
    Reply
    Giải đáp:
    a) 2
    b) 3
    c) 1/49
    Lời giải và giải thích chi tiết:
    a)
    {4^10 . 8^7}/{2^40}
    = {(2^2)^10 . (2^3)^7}/{2^40}
    = {2^20 . 2^21}/{2^40}
    = {2^{20+ 21}}/{2^40}
    = {2^41}/{2^40}
    = 2/1
    = 2
    ————————————-
    b)
    {9^10 . 27^7}/{3^40}
    = {(3^2)^10 . (3^3)^7}/{3^40}
    = {3^20 . 3^21}/{3^40}
    = {3^{20 + 21}}/{3^40}
    = {3^41}/{3^40}
    = 3/1
    = 3
    ————————————-
    c)
    {8^8 . 49^6}/{14^14 . 2^10}
    = {(2^3)^8 . (7^2)^6}/{(2 . 7)^14 . 2^10}
    = {2^24 . 7^12}/{7^14 . 2^14 . 2^10}
    = {2^24 . 7^12}/{7^14 . 2^24}
    = {7^12}/{7^14}
    = 1/{7^2}
    = 1/49

  2. maianhtuanh
    maianhtuanh
    19 Tháng Hai, 2022 at 11:05 sáng
    Reply
    $a)$ $\dfrac{4^{10}\cdot \:8^7}{2^{40}}$
    $=\dfrac{2^{20}\cdot \:8^7}{2^{40}}$
    $=\dfrac{8^7}{2^{20}}$ $=\dfrac{2^{21}}{2^{20}}$
    $=2^{21-20}$ $=2^1$
    $=2$
    $b)$ $\dfrac{9^{10}\cdot \:27^7}{3^{40}}$
    $=\dfrac{3^{20}\cdot \:27^7}{3^{40}}$
    $=\dfrac{27^7}{3^{20}}$ $=\dfrac{3^{21}}{3^{20}}$
    $=3^{21-20}$ $=3^1$
    $=3$
    $c)$ $\dfrac{8^8\cdot \:49^6}{14^{14}\cdot \:2^{10}}$
    $=\dfrac{2^{24}\cdot \:49^6}{14^{14}\cdot \:2^{10}}$ 
    $=\dfrac{2^{24}\cdot \:49^6}{2^{14}\cdot \:7^{14}\cdot \:2^{10}}$ $=\dfrac{2^{10}\cdot \:49^6}{7^{14}\cdot \:2^{10}}$
    $=\dfrac{49^6}{7^{14}}$ $=\dfrac{7^{12}}{7^{14}}$
    $=\dfrac{1}{7^{14-12}}$ $=\dfrac{1}{7^2}$
    $=\dfrac{1}{49}$
     

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222-9+11+12:2*14+14 = ? ( )

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