\(\frac 1 k -\frac 1 {k+1}=\frac {k+1-k}{k.(k+1)}=\frac 1 {k.(k+1)}\)
Theo đề, ta có:
\(\frac 8 9-\frac 1 {72}-\frac 1 {56}-\frac 1 {42}-\frac 1 {30}-\frac 1 {20}-\frac 1 {12}-\frac 1 6-\frac 1 {2}\)
\(=\frac 8 9-(\frac 1 {2}+\frac 1 6+\frac 1 {12}+\frac 1 {20}+\frac 1 {30}+\frac 1 {42}+\frac 1 {56}+\frac 1 {72})\)
\(=\frac 8 9-(\frac 1 {1.2}+\frac 1 {2.3}+\frac 1 {3.4}+\frac 1 {4.5}+\frac 1 {5.6}+\frac 1 {6.7}+\frac 1 {7.8}+\frac 1 {8.9})\)
\(=\frac 8 9-(\frac 1 {1}-\frac 1 2+\frac 1 {2}-\frac 1 3+\frac 1 {3}-\frac 1 4+\frac 1 {4}-\frac 1 5+\frac 1 {5}-\frac 1 6+\frac 1 {6}-\frac 1 7+\frac 1 {7}-\frac 1 8+\frac 1 {8}-\frac 1 9)\)
\(=\frac 8 9-(\frac 1 {1}-\frac 1 9)\)
\(=\frac 8 9-\frac 8 9=0\)
Chúc bạn học tốt!
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