\(ab=bc=ca\Rightarrow\frac{ab}{abc}=\frac{bc}{abc}=\frac{ca}{abc}\)

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

\(\Rightarrow M=\left(\frac{a}{b}\right)^{2016}-\left(\frac{c}{a}\right)^{2017}\)

\(=\left(\frac{a}{a}\right)^{2016}-\left(\frac{a}{a}\right)^{2017}\)

\(=1^{2016}-1^{2017}\)

\(=1-1=0\)

                                  Giải

Ta có: \(a^b=b^c=c^a\)

\(\Leftrightarrow a=b=c\)

\(\Leftrightarrow M=1^{2016}-1^{2017}\)

\(\Leftrightarrow M=1-1\)

\(\Leftrightarrow M=0\)

theo bài ra ta có:

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

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

=> \(\frac{abc}{ca+cb}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)

vì a,b,c khác 0 => ca+cb = ab+ac = bc+ba

=> a = b = c

ta có:

\(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

vậy M = 1

Mik ko bk đúng hay sai đâu nha!

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{a}{a+b}\cdot b=\frac{c}{b+c}\cdot b\)

\(\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\Rightarrow ab+ac=ac+bc\Rightarrow ab=bc\Rightarrow a=c\left(1\right)\)

\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{b}{a+b}\cdot a=\frac{c}{a+c}\cdot a\)

\(\Rightarrow\frac{b}{a+b}=\frac{c}{a+c}\Rightarrow b\left(a+c\right)=c\left(a+b\right)\Rightarrow ab+bc=ac+bc\Rightarrow ab=ac\Rightarrow b=c\left(2\right)\)

\(\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{b}{b+c}\cdot c=\frac{a}{a+c}\cdot c\)

\(\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Rightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow a=b\left(3\right)\)

từ \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow a=b=c\)

\(\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)

Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\iff\)\(\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)

\(\iff\) \(ac+bc=ab+ac=bc+ba\)

+)\(ac+bc=ab+ac\) 

\(\implies\)\(bc=ab\)

\(\implies\) \(c=a\left(1\right)\)

+)\(ab+ac=bc+ba\)

\(\implies\) \(ac=bc\)

\(\implies\) \(a=b\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\implies\) \(a=b=c\)

\(\implies\) \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)

Vậy \(M=1\)

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

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

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

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

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

\(\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a\cdot a+a\cdot a+a\cdot a}{a^2+a^2+a^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Nguyễn Châu Tuấn Kiệt ông có thể giúp tui bài này đc ko

bài này tôi đăng lên rroif mà chẳng ai bít mà trả lời

Ta có \(a^b=b^c=c^a\)

\(\Rightarrow a=b=c\)

\(\Rightarrow M=1^{2016}-1^{2017}\)

\(M=0\)

Nguyễn Châu Tuấn Kiệt ko bt có đúng k??