Ta có : \(\frac{2x-1}{2X+3}\)=\(\frac{2x+3-3-1}{2x+3}\)
=1-\(\frac{4}{2x+3}\)
Muốn \(\frac{2x-1}{2x+3}\)
là số nguyên thì suy ra \(\frac{4}{2x+3}\) nguyên
suy ra 2 x+3 thuộc ư(4)= +-1 , +- 2 , +-4
Nên: 2 x+3=1 suy ra x=-1
2 x+3 = - 1 suy ra x = -2
2x+3= 2 suy ra x =\(\frac{-1}{5}\)
2x+3= -2 suy ra x =\(\frac{-5}{2}\)
2x+3= -4 suy ra x= \(\frac{-4}{2}\)
2x+3 = 4 suy ra x= \(\frac{1}{2}\)
Vậy : x = -1 và x = -2
de phan so nay la so nguyen thi
2x-1 chia het cho 2x+3
ta co 2x-1= 2x+3-4
dan den 4 chia het cho 2x+3
ma uoc cua 4 la 1;4;2;-1;-2;-4
khi 2x+3=1 thi x=-1 khi 2x+3 =-1 thi x=-2
khi 2x+3=4 thi x loai khi 2x+3=-4 thi x loai
khi 2x+3 =-2thi xloai khi 2x+3 =2 thi xloai
vay x=-1 hoac =-2
