Cho a.b.c=1.CM
\(\frac{a}{a.b+a+1}+\frac{b}{bc+c+1}+\frac{c}{ca+c+1}=1\)
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\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{abc.a+abc+ab}\)
Thay abc = 1, ta có:
\(\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)
\(=\frac{ab+a+1}{ab+a+1}\)
\(=1\)
\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{c}{c\left(1+a+ab\right)}+\frac{ac}{ac\left(1+b+bc\right)}+\frac{1}{1+c+ca}\)
\(=\frac{c}{c+ac+abc}+\frac{ac}{ac+abc+abc^2}+\frac{1}{1+c+ca}\)
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
\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)\(=\frac{1}{ab+a+1}+\frac{a}{a\left(bc+b+1\right)}+\frac{abc}{ca+c+abc}\)
\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1\)
Theo bài ra ta có: a.b.c = 1
=> a=1;b=1;c=1
Ta có: A = \(\frac{1}{a.b+a+1}\)\(+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)\(=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}\)\(+\frac{1}{1.1+1+1}\)
\(=\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\)\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1\)
Vậy A = 1
Cho các số a,b,c thỏa mã a.b.c = 1
Tính A = \(\frac{1}{a.b+a+1}+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)
\(A=\)\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
\(=\frac{c}{\left(ab+a+1\right)c}+\frac{ac}{\left(bc+b+1\right).ac}+\frac{1}{ca+c+1}\)
\(=\frac{c}{abc+ac+c}+\frac{ac}{abc^2+abc+ac}+\frac{1}{ca+c+1}\)
\(=\frac{c}{1+ac+c}+\frac{ac}{c+1+ac}+\frac{1}{ca+c+1}\)
\(=\frac{c+ac+1}{1+ac+c}=1\)
\(B=\frac{1}{1+a+ab}+\frac{a}{a+ab+abc}+\frac{abc}{abc+c+ca}\)
\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{abc}{c\left(ab+1+a\right)}\)
\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)
\(=\frac{1+a+ab}{1+a+ab}=1\)