Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\a=b\end{matrix}\right.\) \(\Rightarrow a=b=c\)
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:
\(\Rightarrow\dfrac{abc}{ab+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
\(\Rightarrow ac+bc=ab+ac=bc+ab\)
\(\Rightarrow M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
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
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=1\)
ta có \(\left\{{}\begin{matrix}\dfrac{ab}{a+b}=\dfrac{ac}{a+c}\\\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a.\dfrac{b}{a+b}=a.\dfrac{c}{c+a}\\b.\dfrac{a}{a+b}=b.\dfrac{c}{b+c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a+b}=\dfrac{c}{c+a}\\\dfrac{a}{a+b}=\dfrac{c}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1+\dfrac{b}{a}=1+\dfrac{c}{a}\\1+\dfrac{a}{b}=1+\dfrac{c}{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{a}=\dfrac{c}{a}\\\dfrac{a}{b}=\dfrac{c}{b}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)
Câu 2:
Theo đề, ta có: \(\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ab+10ac+b^2+bc=10ab+10b^2+ac+bc
=>9ac-9b^2=0
=>ac-b^2=0
=>ac=b^2
=>a/b=b/c
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
=> 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:
\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)