Giải thích các bước giải:

Ta có :

$\dfrac{xy+2y+1}{xy+x+y+1}=\dfrac{y(x+1)+y+1}{(x+1)(y+1)}=\dfrac{y}{y+1}+\dfrac{1}{x+1}$

 Chứng minh tương tự ta có :

$\dfrac{yz+2z+1}{yz+y+z+1}=\dfrac{z}{z+1}+\dfrac{1}{y+1}$

$\dfrac{zx+2x+1}{zx+z+x+1}=\dfrac{x}{x+1}+\dfrac{1}{z+1}$

$\rightarrow A=\dfrac{y}{y+1}+\dfrac{1}{x+1}+\dfrac{z}{z+1}+\dfrac{1}{y+1}+\dfrac{x}{x+1}+\dfrac{1}{z+1}$

$\rightarrow A=\dfrac{y+1}{y+1}+\dfrac{x+1}{x+1}+\dfrac{z+1}{z+1}$

$\rightarrow A=3$