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Ta có \(\dfrac{ab}{a+b}\)=\(\dfrac{bc}{b+c}\)=\(\dfrac{ca}{c+a}\)
\(=>\)\(\dfrac{a+b}{ab}\)=\(\dfrac{b+c}{bc}\)=\(\dfrac{c+a}{ca}\)
\(=>\)\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)=\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=\(\dfrac{1}{c}\)+\(\dfrac{1}{a}\)
\(=>\)\(\dfrac{1}{b}\)+\(\dfrac{1}{a}\)=\(\dfrac{1}{c}\)+\(\dfrac{1}{b}\)
\(\dfrac{1}{c}\)+\(\dfrac{1}{b}\)=\(\dfrac{1}{a}\)+\(\dfrac{1}{c}\)
\(\dfrac{1}{a}\)+\(\dfrac{1}{c}\)=\(\dfrac{1}{b}\)+\(\dfrac{1}{a}\)
\(=>\)\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)
\(=>\)a=b=c
Vậy: M=\(\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}\)
= 1
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\a=b\end{matrix}\right.\) \(\Rightarrow a=b=c\)
Thay vào M ta được:
\(M=\dfrac{ab+bc+ac}{a^2+b^2+c^2}=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
theo đề bài ta có:
\(\Rightarrow\dfrac{abc}{ab+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
\(\Rightarrow ac+bc=ab+ac=bc+ab\)
\(\Rightarrow M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
=> ac + bc = ab + ac = bc + ab (do abc \(\ne0\))
=> ac + bc - ab - ac = 0
=> bc - ab = 0
=> b(c - a) = 0
Mà b \(\ne0\) nên c - a = 0 => c = a
Tương tự ta có: a = b
Từ đó có: a = b = c
Thay vào M được:
\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
\(\hept{\begin{cases}\frac{ab}{a+b}=\frac{bc}{b+c}\Rightarrow ab.\left(b+c\right)=\left(a+b\right).bc\Rightarrow abb+abc=abc+bbc\Rightarrow a=c\\\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\left(c+a\right).bc=\left(b+c\right).ca\Rightarrow bcc+abc=abc+cca\Rightarrow a=b\end{cases}\Rightarrow a=b=c}\)
\(M=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
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:
\(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
\(\hept{\begin{cases}\frac{ab}{a+b}=\frac{bc}{b+c}\Rightarrow ab.\left(b+c\right)=\left(a+b\right).bc=ab^2+abc=abc+b^2c\\\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\left(a+c\right).bc=\left(b+c\right).ac\Rightarrow abc=c^2a=abc+c^2b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=c\\a=b\end{cases}\Rightarrow a=b=c\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1}\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=1\)
ta có \(\left\{{}\begin{matrix}\dfrac{ab}{a+b}=\dfrac{ac}{a+c}\\\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a.\dfrac{b}{a+b}=a.\dfrac{c}{c+a}\\b.\dfrac{a}{a+b}=b.\dfrac{c}{b+c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a+b}=\dfrac{c}{c+a}\\\dfrac{a}{a+b}=\dfrac{c}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1+\dfrac{b}{a}=1+\dfrac{c}{a}\\1+\dfrac{a}{b}=1+\dfrac{c}{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a}=\dfrac{c}{a}\\\dfrac{a}{b}=\dfrac{c}{b}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)
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
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)