a, Ta có: \(\frac{a}{a+b}>\frac{a}{a+b+c};\frac{b}{b+c}>\frac{b}{a+b+c};\frac{c}{c+a}>\frac{c}{a+b+c}\)

\(\Rightarrow M>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\left(1\right)\)

b, Ta có: \(\frac{a}{a+b}< \frac{a+b}{a+b+c};\frac{b}{b+c}< \frac{b+c}{a+b+c};\frac{c}{c+a}< \frac{c+a}{a+b+c}\)

\(\Rightarrow M< \frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{c+a}{a+b+c}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\)

Từ (1) và (2) => 1<M<2 hay M không phải là số nguyên

Bạn tham khảo nhé 

\(b)\) Ta có : 

\(\frac{a}{a+b}>\frac{a}{a+b+c}\)

\(\frac{b}{b+c}>\frac{b}{a+b+c}\)

\(\frac{c}{c+a}>\frac{c}{a+b+c}\)

\(\Rightarrow\)\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\)\(M>1\)\(\left(1\right)\)

Lại có : 

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

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

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

\(\Rightarrow\)\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\)\(M< 2\)\(\left(2\right)\)

Từ (1) và (2) suy ra : \(1< M< 2\)

Vậy M không phải là số nguyên 

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

cm: \(1< M< 2\) sẽ thỏa mãn cả a và b

Ta có:

\(M>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\)

vì \(a;b;c>0\Leftrightarrow\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}< 1\)

\(\Rightarrow M< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}=2\)

hay: \(1< M< 2\)

Ta có:\(\frac{a}{b}< \frac{a+m}{b+m}\Rightarrow a\left(b+m\right)< b\left(a+m\right)\)

\(\Leftrightarrow ab+am< ba+bm\Rightarrow am< bm\)

\(\Rightarrow a< b\)(đúng vì  \(\frac{a}{b}< 1\))

D

d nha

\(\frac{a}{b}>1\Rightarrow a>b\)

Ta có :

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

<=> \(b\left(a+m\right)< a\left(b+m\right)\)

<=> \(ab+bm< ab+am\)

<=> \(bm< am\)

<=> \(b< a\)  (Đúng do giả thiết cho)

Vậy ......

Ta có: \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b^2+bm}\)

           \(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b^2+bm}\)

\(\Rightarrow\frac{a}{b}>1\Rightarrow a>b\)

\(\Rightarrow ab+am>ab+bm\)

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

c)

áp dụng BĐT cô si cho 2 số không âm ta có

\(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\)

tương tự

\(b^2+c^2\ge2bc\)

\(c^2+a^2\ge2ac\)

cộng các vế với nhau ta đc

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

=>\(a^2+b^2+c^2\ge ab+bc+ac\) (đpcm)

1.

Gọi vận tốc xe máy là: x (km/h) x >0

Vận tốc ô tô là: x +20 (km/h)

Thời gian xe máy đi là: 10h30 - 6h = 4h30=\(\dfrac{9}{2}h\)

Thời gian đi của ô tô là: 10h30 - 7h30 = 3h

Quãng đường đi của xe máy là: \(\dfrac{9x}{2}\) km

Quãng đường đi của ô tô là: \(\left(x+20\right).3=3x+60\)

Theo đề ra ta có pt:

\(\dfrac{9x}{2}=3x+60\)

\(\Leftrightarrow6x+120=9x\)

\(\Leftrightarrow x=40\) ( nhận)

Vậy vận tốc của xe máy là: 40 km/h

Vận tốc của ô tô là: 40+20 = 60 km/h

Lời giải:
a. 

$\frac{a}{b}<1\Rightarrow a< b\Rightarrow a-b<0$

Xét hiệu $\frac{a}{b}-\frac{a+m}{b+m}=\frac{am-bm}{b(b+m)}=\frac{m(a-b)}{b(b+m)}<0$ do $a-b<0$ và $a,b,m$ là số tự nhiên $>0$

$\Rightarrow \frac{a}{b}<\frac{a+m}{b+m}$

b.

$\frac{a}{b}>1\Rightarrow a> b\Rightarrow a-b>0$

Xét hiệu $\frac{a}{b}-\frac{a+m}{b+m}=\frac{am-bm}{b(b+m)}=\frac{m(a-b)}{b(b+m)}>0$ do $a-b>0$ và $a,b,m$ là số tự nhiên $>0$

$\Rightarrow \frac{a}{b}>\frac{a+m}{b+m}$