a . 3.15 – 4^2 . 1^2019 + 5^2 b. ( 2^78 + 2^79+ 2^80) : ( 2^75+2^76+2^77) c. ( 12x – 140) . 5^14 = 4. 5^10

a . 3.15 – 4^2 . 1^2019 + 5^2
b. ( 2^78 + 2^79+ 2^80) : ( 2^75+2^76+2^77)
c. ( 12x – 140) . 5^14 = 4. 5^10

0 bình luận về “a . 3.15 – 4^2 . 1^2019 + 5^2 b. ( 2^78 + 2^79+ 2^80) : ( 2^75+2^76+2^77) c. ( 12x – 140) . 5^14 = 4. 5^10”

  1. Đáp án:

     

    Giải thích các bước giải:

    a . `3.15 – 4^2 . 1^2019 + 5^2`

    `=45-16.1+25`

    `=29+25`

    `=54`

    b. `( 2^78 + 2^79+ 2^80) : ( 2^75+2^76+2^77)`

    `= (2^78 + 2^78. 2^1 + 2^78. 2^2) : (2^75 + 2^75 . 2^1 + 2^75. 2^2)`

    `= [2^78. (1 + 2^1 + 2^2)] : [2^75. (1 + 2^1 + 2^2)]`

    `= 2^78 : 2^75`

    `= 2^3`

    `= 8`

    c. `( 12x – 140) . 5^14 = 4. 5^10`

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  2.  a.  3 . 15 – $4^{2}$ . $1^{2019}$ + $5^{2}$ 

        = 45 – 16 . 1 + 25

        = 45 – 16 + 25

        = 29 + 25

        = 54  

     b.  ( $2^{78}$ + $2^{79}$ + $2^{80}$ ) ÷ ( $2^{75}$ + $2^{76}$ + $2^{77}$ ) 

        = (  $2^{78}$  + $2^{78}$ . 2 + $2^{78}$ . $2^{2}$ ) + ( $2^{75}$ + $2^{75}$ . 2 + $2^{75}$ . $2^{2}$ ) 

        = [ $2^{78}$ . ( 1 + 2 + $2^{2}$ ] + [ $2^{75}$ . ( 1 + 2 + $2^{2}$ ) ] 

        = $2^{78}$ ÷  $2^{75}$ 

        = $2^{3}$

        = 8

    Ý c số to quá b ạ 

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