Đề bài sai nhé, chỗ \(\frac{1}{b.c+b+1}\) phải là \(\frac{b}{b.c+b+1}\) ms đúng
Ta có:
\(\frac{1}{a.b+a+1}+\frac{b}{b.c+b+1}+\frac{1}{a.b.c+b.c+b}=\frac{a.b.c}{a.b+a+a.b.c}+\frac{b}{b.c+b+1}+\frac{1}{1+b.c+b}\)
\(=\frac{a.b.c}{a.\left(b+1+b.c\right)}+\frac{b}{1+b.c+b}+\frac{1}{1+b.c+b}\)
\(=\frac{b.c}{b+1+b.c}+\frac{b}{1+b.c+b}+\frac{1}{1+b.c+b}=\frac{b.c+b+1}{1+b.c+b}=1\left(đpcm\right)\)