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