tìm a thuộc số tự nhiên để a+1 là bội của a-1
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(=) 52x-3 -2 =3
(=) 52x-3 = 5
(=)2x-3=1
(=)2x=4
(=)x=2
#Học-tốt
a)trên tia Ox có ON <OM (7cm<15cm)
nên điểm n nằm giữa O,M (1)
b)từ (1) suy ra :
ON+NM=OM ( ON=7cm,Om=15cm)
hay 7+MN=15
MN=15-7
MN=8cm
c)trên tia Ox có MI < MN (1cm<8cm)
nên I nằm giữa 2 điểm M,N (2)
nên: NI+IN=NM
hay NI+1=8
NI=8-1
NI=7cm
mà ON=7cm
suy ra ON=NI(=7cm) (3)
từ (1) (2) cho ta :
N nằm giữa 2 điểm O,I (4)
từ (3) (4) ta đc
N là trung điểm của OI
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| x - 6 | = 89
ta có 2 trường hợp
th1
x -6 = 89
x = 89 + 6
x = 95
th2
x -6 = -89
x = (-89) + 6
x = -82
=> x = 95;-82
giáng sinh vv
Merry Christmas
a + 1 là bội của a - 1
=> a + 1 chia hết cho a - 1
=> [(a - 1) + 2] chia hết cho a - 1
=> 2 chia hết cho a - 1, a - 1 thuộc ước của 2 = {1; 2}
a - 1 = 1
=> a = 2
a - 1 = 2
=> a = 3
Vậy a thuộc {2; 3}