Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\hept{\begin{cases}\frac{ab}{a+b}=\frac{bc}{b+c}\Rightarrow ab.\left(b+c\right)=\left(a+b\right).bc\Rightarrow abb+abc=abc+bbc\Rightarrow a=c\\\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\left(c+a\right).bc=\left(b+c\right).ca\Rightarrow bcc+abc=abc+cca\Rightarrow a=b\end{cases}\Rightarrow a=b=c}\)
\(M=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
p/s: bài này có nhiều cách lắm, cách này ko đc thì thử làm cách khác =))
\(\frac{ab}{a+b}=\frac{bc}{b+c}\Rightarrow ab\left(b+c\right)=\left(a+b\right)bc\)
\(\Rightarrow ab^2+abc=abc+b^2c\Rightarrow ab^2=b^2c\Rightarrow a=c\) (1)
\(\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow bc\left(c+a\right)=\left(b+c\right)ca\)
\(\Rightarrow bc^2+bca=bca+c^2a\Rightarrow bc^2=c^2a\Rightarrow b=a\)(2)
Từ (1) và (2) được a = b = c
Khi đó:
\(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{a}{a+b}\cdot b=\frac{c}{b+c}\cdot b\)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\Rightarrow ab+ac=ac+bc\Rightarrow ab=bc\Rightarrow a=c\left(1\right)\)
\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{b}{a+b}\cdot a=\frac{c}{a+c}\cdot a\)
\(\Rightarrow\frac{b}{a+b}=\frac{c}{a+c}\Rightarrow b\left(a+c\right)=c\left(a+b\right)\Rightarrow ab+bc=ac+bc\Rightarrow ab=ac\Rightarrow b=c\left(2\right)\)
\(\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{b}{b+c}\cdot c=\frac{a}{a+c}\cdot c\)
\(\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Rightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow a=b\left(3\right)\)
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 \(M=1\)
Nếu ab là ab thì mk giải thế này:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Leftrightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}=\frac{10c+a}{c+a}\)
Theo t/c dãy tỉ số=nhau:
\(\frac{10a+b}{a+b}=\frac{10b+c}{b+c}=\frac{10c+a}{c+a}=\frac{\left(10a+b\right)+\left(10b+c\right)+\left(10c+a\right)}{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\)
\(=\frac{\left(10a+a\right)+\left(10b+b\right)+ \left(10c+c\right)}{\left(a+a\right)+\left(b+b\right)+\left(c+c\right)}=\frac{11a+11b+11c}{2a+2b+2c}=\frac{11\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{11}{2}\)
do đó: \(\frac{10a+b}{a+b}=\frac{11}{2}\Rightarrow\left(10a+b\right).2=11.\left(a+b\right)\Rightarrow20a+2b=11a+11b\)
\(\Rightarrow20a-11a=11b-2b\Rightarrow9a=9b\Rightarrow a=b\)
Tương tự với b=c;c=a
=>\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0^3+0^3+0^3=0\)
Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a\cdot a+a\cdot a+a\cdot a}{a^2+a^2+a^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)