Question

Chứng minh rằng:

\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\) biết abc=1

Answer 1

Thay abc = 1 vào biểu thức ta có

\(\frac{a.abc}{ab+abc.a+abc}+\frac{b}{bc+b.acb+abc}+\frac{c}{ac+c+1}\)

= \(\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+ab^2c+abc}+\frac{c}{ac+c+1}\)

= \(\frac{a^2bc}{ab\left(ac+c+1\right)}+\frac{b}{b\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

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

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

= 1 (đpcm)

Nếu có gì không hiểu nhớ nt cho mình nha

Answer 2

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

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

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

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