Đáp án:
Bài `2`
`b,`
`(2^3 + 4^2 + 76) - 2 (x + 3) = 5^2 . 2`
`-> (8 + 16 + 76) - 2x - 6 = 25 . 2`
`-> 100 - 2x - 6 =50`
`-> 94 - 2x = 50`
`-> 2x = 44`
`-> x = 22`
Vậy `x=22`
`c,`
`16^x : 64 = 4^5`
`-> (4^2)^x : 4^3 = 4^5`
`-> 4^{2x} = 4^5 . 4^3`
`-> 4^{2x} = 4^8`
`-> 2x = 8`
`-> x = 4`
Vậy `x=4`
`d,`
`2^x + 2^{x + 1} + 2^{x + 2} = 960 - 2^{x + 3}`
`-> 2^x + 2^{x + 1} + 2^{x + 2} + 2^{x + 3} = 960`
`-> 2^x + 2^x . 2 + 2^x . 2^2 + 2^x . 2^3 = 960`
`-> 2^x . (1 + 2 + 2^2 + 2^3) = 960`
`-> 2^x . (1 + 2 + 4 + 8) = 960`
`-> 2^x . 15 = 960`
`-> 2^x=64`
`-> 2^x= 2^6`
`->x=6`
Vậy `x=6`
$\\$
Bài `3`
`(x + 1) (2y - 5) = 143`
`-> (x + 1) (2y - 5) =1 . 143 = 143 . 1 = 11 . 13 = 13 . 11` (Thỏa mãn `x,y ∈ NN`)
Ta có bảng sau :
$\begin{array}{|c|c|c|c|c|c|c|}\hline x +1& 1& 143 & 11 & 13 \\\hline 2y-5& 143 & 1 & 13 & 11 \\\hline x &0 &142 &10 &12 \\\hline y &74 &3 &9 &8 \\\hline\end{array}$
Vậy cặp `(x;y)` thỏa mãn là : `(0;74), (142;3), (10;9), (12;8)`
