\(x< y\)

\(\Rightarrow\frac{a}{m}< \frac{b}{m};m>0\)

\(\Rightarrow a< b\)

\(\Rightarrow\frac{a+a}{m}< \frac{a+b}{m}\)

\(\Rightarrow\frac{a+a}{2m}< \frac{a+b}{2m}\)

\(\Rightarrow\frac{2a}{2m}< \frac{a+b}{2m}\)

\(\Rightarrow\frac{a}{m}< \frac{a+b}{2m}\)

\(\Rightarrow x< z\left(1\right)\)

Tương tự lại có :

\(\frac{a+b}{m}< \frac{b+b}{m}\)

\(\Rightarrow\frac{a+b}{2m}< \frac{b+b}{2m}\)

\(\Rightarrow\frac{a+b}{2m}< \frac{2b}{2m}\)

\(\Rightarrow\frac{a+b}{2m}< \frac{b}{m}\)

\(\Rightarrow z< y\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow x< z< y\)

Vậy \(x< z< y.\)

x=a/m<y=b/m=>a<b

=>x=2a/2m<y=2b/2m

2a<a+b =>x=2a/2m<z=a+b/2m

a+b<2b =>z=a+b/2m<2b/2m

=>đpcm

trong sgk toán 7 có, mà nó hướng dẫn rồi thây

t

Ta có: 

 x = \(\frac{a}{m}=\frac{a+a}{2m}\)

y = \(\frac{b}{m}=\frac{b+b}{2m}\)

Vì x < y, \(\Rightarrow\)a < b

Vì a < b, \(\Rightarrow\)\(\frac{a+a}{2m}\) <  \(\frac{a+b}{2m}\) <  \(\frac{b+b}{2m}\)

Vậy x < z < y nếu z = \(\frac{a+b}{2m}\)

Do x < y

=> a/m < b/m

=> a/m + a/m < a/m + b/m < b/m + b/m

=> 2a/m < a+b/m < 2b/m

=> a/m < a+b/m : 2 < b/m

=> a/m < a+b/m × 1/2 < b/m

=> a/m < a+b/2m < b/m

=> x < z < y

=> am<bm

=>am+am<am+bm =>a.2m<m.(a+b)

=>a/m<a+b/2m         (1)

=>am+bm<bm+bm=>m(a+b)<b.2m

=>a+b/2m<b/m      (2)

tu (1) va (2)

=>a/m<a+b/m2<b/m

Theo đề ta có :

\(x=\frac{a}{m}\) \(;\)\(y=\frac{b}{m}\)

mà \(x< y\) \(\Rightarrow\frac{a}{m}< \frac{b}{m}\Rightarrow a< b\)

Có : \(x=\frac{a}{m}\Rightarrow x=\frac{2a}{2m}=\frac{a+a}{2m}\)     ; \(z=\frac{a+b}{2m}\)    và \(y=\frac{b}{m}\Rightarrow y=\frac{2b}{2m}=\frac{b+b}{2m}\) 

* Vì a < b \(\Rightarrow\) a+a < a+b \(\Rightarrow\frac{a+a}{2m}< \frac{a+b}{2m}\)\(\Rightarrow x< z\) \(\left(1\right)\)

* Vì \(a< b\)\(\Rightarrow a+b< b+b\Rightarrow\frac{a+b}{2m}< \frac{b+b}{2m}\Rightarrow z< y\)\(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) nên ta có :

\(\frac{a+a}{2m}< \frac{a+b}{2m}< \frac{b+b}{2m}\Rightarrow x< z< y\) \(\left(đpcm\right)\)

Vì x < y (\(\frac{a}{m}< \frac{b}{m}\)) và m > 0 nên a < b .

 x = \(\frac{a}{m}=\frac{2a}{2m}\); y = \(\frac{b}{m}=\frac{2b}{2m}\); z = \(\frac{a+b}{2m}\). Ta có :

a < b nên a + a < a + b < b + b hay 2a < a + b < 2b => \(\frac{2a}{2m}< \frac{a+b}{2m}< \frac{2b}{2m}\)=> x < z < y

Vì x<y nên a<b. Ta có \(x=\frac{a}{m}=\frac{2a}{2m},y=\frac{b}{m}=\frac{2b}{2m}\)

Chọn \(z=\frac{2a+1}{2m}\).Do 2a<2a+1 nên x<z(1)

Do a<b nên a+1 < b suy ra 2a+1< 2b

TA có 2a+1< 2a+2< 2b nên 2a+1<2b do đó z<y(2)

Từ (1),(2) suy ra x<z<y

Ta có: x<y => \(\frac{a}{m}< \frac{b}{m}\)<=> a<b

Lại có:\(x=\frac{a}{m}=\frac{2a}{2m};y=\frac{b}{m}=\frac{2b}{2m}\)

vì a<b (a, b thuộc Z) <=> a+1 =< b hay 2a+2 =< 2b

=> 2a <2a+1<2a+2=<2b hay 2a<2a+1<2b

do đó: \(\frac{2a}{2m}< \frac{2+1}{2m}< \frac{2b}{2m}\)

=> x<y<z

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