cac ban khong lam thi minh lam nhe
sang tien cho ****
he he he he!
Vi :\(0<\frac{a}{b}<1\left(b>0\right)\) nen a<b ma m>0, do do am<bm , them ab vao 2 ve :
ab+am<ab+bm hay a(b+m)<b(a+m) ma b>0 va b+m>0 nen suy ra :
\(\frac{a}{b}<\frac{a+m}{b+m}\)
**** nhe moi ng
G/s x = \(\frac{a}{m}\) , y = \(\frac{b}{m}\) ( a, b, m \(\in Z\) , m > 0 ) , x < y . Hay chung to rang neu chon z= \(\frac{a+b}{2m}\) thi ta co x < z < y .
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
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=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 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
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 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 la so nguyen, m>0) va x<y. hay chung to rang neu chon z=a+b/2m thi ta co x<z<y
Chung minh rang neu a^2=bc thi
a) \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)
b) \(\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\)