Question

cho: abc = 1

Chứng minh rằng: \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)

giúp với nha

Answer 1

Answer 2

\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

=\(\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ac+c+abc}\)

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

=\(\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{1}{a+1+ab}\)

=\(\dfrac{a+1}{ab+a+1}+\dfrac{ab}{abc+ab+a}\)

=\(\dfrac{a+abc}{ab+a+abc}+\dfrac{ab}{abc+ab+a}\)

=\(\dfrac{a+abc+ab}{ab+a+abc}=1\) (đpcm)

Answer 3

thui mk làm đc rồi ko cần giúp nữa