Bài 2 : Theo ví dụ trên ta có : \(\frac{a}{b}< \frac{c}{d}\)=> ad < bc

Suy ra :

\(\Leftrightarrow ad+ab< bc+ba\Leftrightarrow a(b+d)< b(a+c)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)

Mặt khác : ad < bc => ad + cd < bc + cd

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

Vậy : ....

b, Theo câu a ta lần lượt có :

\(-\frac{1}{3}< -\frac{1}{4}\Rightarrow-\frac{1}{3}< -\frac{2}{7}< -\frac{1}{4}\)

\(-\frac{1}{3}< -\frac{2}{7}\Rightarrow-\frac{1}{3}< -\frac{3}{10}< -\frac{2}{7}\)

\(-\frac{1}{3}< -\frac{3}{10}\Rightarrow-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}\)

Vậy : \(-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}< -\frac{2}{7}< -\frac{1}{4}\)

Vì a=b suy ra a/b = 1

Từ đó suy ra a+n=b+n

Suy ra a+n/b+n=1

a/b=1=a+n/b+n suy ra a/b=a+n/b+n

TA có 

\(\frac{a}{b}-\frac{a+c}{b+c}=\frac{a\left(b+c\right)}{b\left(b+c\right)}-\frac{b\left(a+c\right)}{b\left(b+c\right)}\)

\(=\frac{ab+ac-ab-bc}{b\left(b+c\right)}=\frac{ac-bc}{b\left(b+c\right)}=\frac{c\left(a-b\right)}{b\left(b+c\right)}\)

vì a>b => a-b > 0 => c(a-b) > 0 

=> \(\frac{c\left(a-b\right)}{b\left(b+c\right)}>0\)

\(=>\frac{a}{b}-\frac{a+c}{b+c}>0\)

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

=> đpcm

b)   Ta có a+b < a+b+c ; b+c < a+b+c ; c+a < a+b+c

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

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

Lại có 

Áp dùng câu a ta có a< a+b ; b< b+c ; c<c+a

=> \(\frac{a}{a+b}< \frac{a+c}{a+b+c};\frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{c+a}< \frac{c+b}{a+b+c}\)

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

Từ (1) và (2) => dpcm

- Cậu ơi, đpcm là cái gì???

thưa chị e chịu !!!

má ơi e rảnh lắm hả e

vãi thế

1. \(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)

A nguyên nên \(3⋮n-2\). Vậy \(n-2\in\left(1,-1,3,-3\right)\Rightarrow n\in\left(3,1,5,-1\right)\)thì A nguyên.

2. a,Ta cần CM  \(\frac{a}{b}< \frac{a+c}{b+c}\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow ab+ac< ab+bc\Rightarrow ac< bc\)(luôn đúng)

Suy ra điều phải chứng minh.

b, Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

Có:(suy ra từ phần a) \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Vậy \(1< \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\)

BẤM ĐÚNG CHO MÌNH, KO THÌ LẦN SAU KO GIÚP NỮA

Để \(A=\frac{n+1}{n-2}\)có giá trị nguyên => n + 1 chia hết cho n-2

\(=>\left(n-2\right)+3⋮\)\(n-2\)

Mà \(\left(n-2\right)⋮\)\(n-2\)

\(=>3⋮\)\(n-2\)

\(=>n-2\inƯ\left(3\right)=\){1;-1;3;-3}

Ta có bảng :

Vậy \(n\in\){3;1;5;-1} để \(A=\frac{n+1}{n-2}\in Z\)

Câu 2 :

a)Vì \(a< b=>\frac{a}{b}< 1\)

Ta so sánh : \(\frac{a}{b}\)và \(\frac{a+c}{b+c}\)

Ta có : \(\frac{a}{b}=\frac{a.\left(b+c\right)}{b.\left(b+c\right)}=\frac{ab+ac}{b.\left(b+c\right)}\)

             \(\frac{a+c}{b+c}=\frac{b.\left(a+c\right)}{b.\left(b+c\right)}=\frac{ab+bc}{b.\left(b+c\right)}\)

Vì a,b,c \(\in\)N* mà a<b => \(ab+ac< ab+bc=>\frac{ab+ac}{b.\left(b+c\right)}< \frac{ab+bc}{b.\left(b+c\right)}=>\frac{a}{b}< \frac{a+c}{b+c}\left(đpcm\right)\)

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

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

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

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

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

Vì a,b,c \(\in\)N* mà a<b => \(\frac{a}{a+b}< 1\)=>\(\frac{a}{a+b}< \frac{a+c}{a+b+c}\)

Tương tự như vậy ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}+\frac{c+b}{a+b+c}=\frac{a+c+b+a+c+b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)

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

 Từ \(\left(1\right);\left(2\right)=>1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}< 2\left(đpcm\right)\)