Toán Tính : `(2^78 + 2^79 + 2^80) : (2^75 + 2^76 + 2^77)` 16/07/2021 By Gabriella Tính : `(2^78 + 2^79 + 2^80) : (2^75 + 2^76 + 2^77)`
`(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` (bớt cả `2` vế đi `(1 + 2^1 + 2^2)`) `= 2^3` `= 8` Trả lời
Đáp án: 8 Giải thích các bước giải: $\dfrac{2^{78}+2^{79}+2^{80}}{2^{75}+2^{76}+2^{77}}$$= \dfrac{2^{78}\left ( 1+2+2^{2} \right )}{2^{75}\left ( 1+2+2^{2} \right )}$$= \dfrac{2^{78}}{2^{75}}= 2^{3}= 8$ Trả lời
`(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` (bớt cả `2` vế đi `(1 + 2^1 + 2^2)`)
`= 2^3`
`= 8`
Đáp án: 8
Giải thích các bước giải:
$\dfrac{2^{78}+2^{79}+2^{80}}{2^{75}+2^{76}+2^{77}}$
$= \dfrac{2^{78}\left ( 1+2+2^{2} \right )}{2^{75}\left ( 1+2+2^{2} \right )}$
$= \dfrac{2^{78}}{2^{75}}= 2^{3}= 8$