Do ƯCLN(a; b)=16 => a = 16.m; b = 16.n [(m;n)=1; (m > n)]
Ta có: 16.m + 16.n = 128
=> 16.(m + n) = 128
=> m + n = 128 : 16 = 8
Mà m > n; (m;n)=1 => m = 7; n = 1 hoặc m = 5; n = 3
+ Với m = 7; n = 1 thì a = 16.7 = 112; b = 16.1 = 16
+ Với m = 5; n = 3 thì a = 16.5 = 80; b = 16.3 = 48
Vậy các cặp số (a;b) thỏa mãn đề bài là: (112;16) ; (80;48)
UCLN (a,b) - 6 nên a = 6a', b = 6b' và UCLN (a,b) = 1.
Theo đề bài a'b' = 63 =3.3.7
Do a > b nên a'>b'.' Chọn 2 số a' và b' có tích = 63, nguyên tố cùng nhau. a' > b' ta được.
| a' | 63 | 9 |
| b' | 1 | 7 |
Do đó.
| a | 387 | 54 |
| b | 6 | 42 |
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ƯCLN(a,b)=16
Đặt a=16x
b=16y
x, y là hai số nguyên tố cùng nhau, x,y thuộc N và x,y khác 0.
Vì a>b =>x>y
ta có: a+ b= 128
<=> (16x)+(16y)= 128
<=> 16. (x+y)= 128
<=>x+y= 128:16=8
Vìx>y, x,y thuộc N sao và x,ylà hai số nguyên tố cùng nhau, ta có các trường hợp sau:
x 7 6 5
y 1 2 3
Nếu x=7,y=1
=> a=16x=16.7=112
b=16y=16.1=16
Nếu x=6,y=2
=> a=16x=16.6=96
b=16y=16.2=32
Nếu x=5,y=3
=>a=16x=16.5=80
b=16y=16.3=48
vậy ....................
