Cho xyz=2 và 2+x+xy[tex]\neq[/tex]0.Tính B=[tex]\frac{1}{1+y+yz}+\frac{2}{2+2z+xz}+\frac{2}{2+x+xy}[/tex]
$B=\dfrac1{1+y+yz}+\dfrac{xyz}{xyz+xyz^2+xz}+\dfrac{xyz}{xyz+x+xy}
\\=\dfrac1{1+y+yz}+\dfrac{xyz}{xz(y+yz+1)}+\dfrac{xyz}{x(yz+1+y)}
\\=\dfrac1{1+y+yz}+\dfrac{y}{y+yz+1}+\dfrac{yz}{yz+1+y}
\\=1$