\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)

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

\(=\frac{c}{c+ac+abc}+\frac{ac}{ac+abc+abc^2}+\frac{1}{1+c+ca}\)

thay a.b.c=1 Ta đc:

\(a=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+a}\) cộng 3 phân số cùng mẫu c+ac+1

\(=\frac{c+ac+1}{c+ac+1}=1\)

tick cho mk vs nhé

Cho a,b,c thuộc R và a.b.c=1.chứng minh  \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\)

Giải:Ta có:\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{a.c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{c}{ca+c+1}\)

\(=\frac{ac}{ac+c+1}+\frac{1}{c+1+ac}+\frac{c}{ca+c+1}\)

\(=\frac{ac+1+c}{ac+c+1}=1\)

Suy ra điều phải chứng minh

\(B=\frac{1}{1+a+ab}+\frac{a}{a+ab+abc}+\frac{abc}{abc+c+ca}\)

\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{abc}{c\left(ab+1+a\right)}\)

\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)

\(=\frac{1+a+ab}{1+a+ab}=1\)

Có:

\\(S=\\frac{1}{1+a+ab}+\\frac{1}{1+b+bc}+\\frac{1}{1+c+ca}\\)

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

\\(S=\\frac{c}{c+ab+abc}+\\frac{ac}{ac+abc+abc^2}+\\frac{1}{1+c+ca}\\)

Thay a.b.c =1 ta được:

\\(S=\\frac{c}{c+ac+1}+\\frac{ac}{ac+1+c}\\frac{1}{1+c+a}\\)(cộng 3 phân số cùng mẫu c+ac+1)

\\(S=\\frac{c+ac+1}{c+ac+1}=1\\)

\\(\\Rightarrow S=1\\)

Ta có \(\left\{{}\begin{matrix}B=\text{​​}\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\\abc=1\end{matrix}\right.\)

\(\Rightarrow B=\frac{1}{1+a+ab}+\frac{a}{a+ab+abc}+\frac{ab}{ab+abc+abca}\)

\(\Rightarrow B=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)

\(\Rightarrow B=\frac{1+a+ab}{1+a+ab}=1\)

Vậy B = 1

@@ Học tốt

Chiyuki Fujito

Tái bút : mà bài này còn lận 5 cách nx cơ