Đáp án:
\(A=1\)
Giải thích các bước giải:
\(A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\\⇒ A=\dfrac{1}{b+1+\dfrac 2a}+\dfrac{b}{bc+b+1}+\dfrac2{a+2+\dfrac2c}\\⇒A=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{abc}{a+abc+ab}\\⇒A=\dfrac{b+1}{bc+b+1}+\dfrac{abc}{a\left(bc+b+1\right)}\\⇒ A=\dfrac{b+1}{bc+b+1}+\dfrac{bc}{bc+b+1}\\⇒ A=\dfrac{bc+b+1}{bc+c+1}\\⇒ A=1\)