xyz=A , theo bài ra ta có:
(x+1)yz=A+1 ; x(y+2)z=A+2 ; xy(z+2)=A+8
Do đó ta có :xyz+yz=A+1
xyz+2xz=A+2
xyz+2xy=A+8
Mà xyz=A nên yz=1 ; 2xz=2 ; 2xy=8.
Do đó, yz=1; xz=1; xy=4
suy ra x=y mà xy=4 nên x=y=2
suy ra z=1/2
Gia su x = a/m;y = b/m (a;b;m thuoc Z;m>0) va x< y. Hay chung to rang neu chon z = a+b/2m thi ta co x<y<z
gia su x=a/m,y=b/m (a,b,m thuoc Z,m>0) va x>y.Hay chung to rang neu chon z=a+b/2m thi ta co x<z<y
gia su x=a/m,y=b/m(a,b,m€z,m>0,x<y)
hay chung to rang neu chon Z=a+b/2•m thi ta co x<z<y
gia su x=a/m. y=b/m (a,b,m la so nguyen, m>0) va x<y. hay chung to rang neu chon z=a+b/2m thi ta co x<z<y
giả su x =a/m , y = b/m (a,b thuoc z, m >0) va x <y. hay chung to rang neu chon z=a+b/2m thi ta co x<z <y
giai gium minh voi
Gia su \(x=\frac{a}{m},y=\frac{b}{m}\)va x<y.Hay chung to rang neu chon \(z=\frac{a+b}{2m}\)thi ta co x<y<z
Su dung tinh chat neu a,b,c thuoc zva a<b thi a+c<b+c
giup mik voi nha tik cho cam on
CMR neu x/a+2b+c=y/2a+b-c=z/4a-4b+c
thi a/x+2y+z=b/2x+y-z=c/4x-4y+z
ai tra loi nhanh va dung minh se tick cho (chi tiet)
gia su x=a/m, y=b/m (a,b,m thuoc z, m>0) va x<y.hay chung to rang neu chon z=a+b/2m thi ta co x<z<y.
minh moi hoc lop 7 minh ko hieu bai nay ai giai giup minh voi
neu a thuoc z thi x=y vay ai tra loi dc cau hoi nay
