Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\) 

==> a=b ; b=c ; c=a ========> a=b=c

Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)
a=b=c=2017

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

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

\(\frac{a}{b}=1\Rightarrow a=b\); \(\frac{b}{c}=1\Rightarrow b=c\); \(\frac{c}{a}=1\Rightarrow c=a\)

Suy ra : a = b = c = 1

Nếu a = 2017 thì : b = c = 2017

A/b=b/c=c/a   va a.b.c khac 0

Ap dung ting chat day ti so bang nhau ta co

A/.........=a+b+c/b+c+a=1

=)a/b=1=)a=b

     b/c=1=)b=c

      Mà a=b,b=c=)a=b=c(1)

 Mà   a=2017(2)

Tù 1và 2=)a=b=c=2017

Vay b=2017,c=2017

a+n/b+n=a/b

vì +n/+n hết chỉ còn a/b thôi nên bằng

CHÚC BẠN HỌC GIỎI

K MÌNH NHÉ

a+n/b+n>a/b

*Nếu a=b thìa+n/b+n=a/b

*Nếu a>b thì a+n/b+n>a/b

*Nêu a<b thì a+n/b+n<a+b

Thêm đk \(a,b,c\ne0\)

Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)

\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)

\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)

\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)

\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)

\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

dit me

Đặt \(\left(\frac{a}{b};\frac{c}{b}\right)=\left(x;y\right)\) ta có \(\frac{1}{x}+\frac{1}{y}=2\)

\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{\frac{a}{b}+1}{\frac{2a}{b}-1}+\frac{\frac{c}{b}+1}{\frac{2c}{b}-1}=\frac{x+1}{2x-1}+\frac{y+1}{2y-1}\)

\(=1+\frac{3}{2}\left(\frac{1}{2x-1}+\frac{1}{2y-1}\right)=1+\frac{3}{2}.\frac{2x+2y-2}{4xy-2\left(x+y\right)+1}=1+3.\frac{x+y-1}{1}\ge4\)

Do \(\frac{1}{x}+\frac{1}{y}=2\Rightarrow x+y\ge2\)

đpcm

Bài 1 :

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\left(1\right)\)

\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)

Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)

Bài 2:

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

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

\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)

\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)

\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)

Chúc bạn học tốt ( -_- )

Bài 1:

ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}< 1\)

\(\Rightarrow A< 1\)(1) 

ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)

                                                                               \(=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)

\(\Rightarrow B>1\)(2)

Từ (1);(2) => A<B

Bài 2:

ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(\Rightarrow S=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(S=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

thay số: \(S=7.\frac{7}{10}-3\)

\(S=4\frac{9}{10}-3\)

\(S=1\frac{9}{10}=\frac{19}{10}\)

mà  \(1\frac{8}{11}=\frac{19}{11}\)

\(\Rightarrow\frac{19}{10}>\frac{19}{11}\)

\(\Rightarrow S>\frac{19}{11}\)

\(\Rightarrow S>1\frac{8}{11}\)

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)

                                               \(=\frac{7}{b+c}-\frac{b+c}{b+c}+\frac{7}{c+a}-\frac{c+a}{c+a}+\frac{7}{a+b}-\frac{a+b}{a+b}\)

                                                \(=\frac{7}{b+c}-1+\frac{7}{c+a}-1+\frac{7}{a+b}-1\)

                                                \(=\frac{7}{b+c}+\frac{7}{c+a}+\frac{7}{a+b}-3\)  

                                                \(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\) \(.Thay\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{7}{10}\)

                                               \(\Rightarrow S=7.\frac{7}{10}-3=\frac{49}{10}-3=1\frac{9}{10}>1\frac{8}{11}\)

                                              Vậy\(S>1\frac{8}{11}\)

S>19/11

100% luôn

\(S>1\frac{8}{11}\)