giải:
ta gọi 2 số đó là a,b
a= 27k
b= 27y
theo đề bài ta có :a x b = 8748 thay vào ta có 27k x 27y = 8748
=729 .k.y = 8748
k.y = 8748 : 729
k.y=12
ta có bảng sau
| k | 1 | 12 | 3 | 4 | 6 | 2 |
| y | 12 | 1 | 4 | 3 | 2 | 6 |
| a = 27k | 27 | 324 | 81 | 108 | 162 | 54 |
| b= 27y | 324 | 27 | 108 | 81 | 54 | 162 |
=> có 4 cặp gồm : a= 1và b=12;a= 12 và b=1 ; a = 3 và b=4 ; a=4 và b =3 ( 6,2 và 2,6 loại vì có ƯCLN là 54)
tic nhé
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Gọi hai số cần tìm là a và b
Vì tích của chúng bằng 8748 nên ta có:
a.b=8748 (1)
Do UCLN(a,b)=27
=> a chia hết cho 27
b chia hết cho 27
=> a=27x
b=27y (a,b)=1 (2)
Từ (1) và (2) suy ra
27x.27y=8748
729.x.y=8748
x.y=12
Do đó ta có bảng sau:
| x | 1 | 12 | 3 | 4 |
| y | 12 | 1 | 4 | 3 |
| a | 27 | 324 | 81 | 108 |
| b | 324 | 27 | 108 | 81 |
Vậy: Các cặp a và b thỏa mãn là: 27 và 324; 324 và 27; 81 và 108; 108 và 81
Hỏi các bạn nè làm sao mà gửi câu hỏi được
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gọi 2 stn cần tìm là a,b(a>b,a,b thuộc N)
theo bài ra ta có:
ucln của ab =27
=>a=27m,b=27n
Với m,n=1 và m,>n
Khi đó a.b=27m.27n=87478
=>m,n thuộc Ư(12) và m,n=1, m>n
ta lập bả
m | n | a | b |
| 4;12 | 3;1 | 108;324 | 81;27 |
Vậy các cặp số cần tìm là (108,81) (324,270)
k mink nhé cô mink bảo thế!
gọi 2 số cần tìm là a,b
a =27m
b =27n
ƯCLN (m,n)=1
Ta có 27m.27n=8784
=> mn =8784:(27.27)=12
do m>n và ƯCLN (m,n)=1
ta có bảng sau:
| m | 5 | 1 |
| n | 7 | 11 |
=>
| m | 135 | 27 |
| n | 189 | 297 |
các bạn tự làm phần đáp số nha!!!!
