Ta có
[tex]\frac{1}{ab+b+1}=\frac{1}{ab+b+abc}=\frac{1}{b(1+a+ac)}=\frac{1}{\frac{1}{ac}(ac+a+1)}=\frac{ac}{ac+a+1}[/tex]
[tex]\frac{1}{bc+c+1}=\frac{1}{bc+c+abc}=\frac{1}{c(b+1+ab)}=\frac{1}{\frac{1}{ab}(b+abc+ab)}=\frac{1}{\frac{1}{a}(1+ac+a)}=\frac{a}{ac+a+1}[/tex]
=> [tex]P=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}=1[/tex]
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