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=1 phần 3*4+1 phần 4*5+1 phần 5*6+...+1 phần 98*99
=1 phần 3-1 phần 4+ 1 phần 4- 1 phần 5+...+1 phần 98-1 phần 99
=1 phần 3- 1 phần 99 =32 phần 99
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{33-1}{99}=\dfrac{32}{99}\)
\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{98\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
1/3.4 + 1/4.5 + ...+1/98.99
= 1/3-1/4+1/4-1/5+...+1/98-1/99
= 1/3-1/99= 32/99
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\\ =\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)
A = 1/1x2 +1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8
=1-1/2+1/2-1/3+1/3-1/4+..+1/7-1/8
=1-1/8
=7/8
Đặt tổng trên là A
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7 = 1-1/7 = 6/7
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
= 1/2 - 1/3 + 1/3 -1/4 + 1/4 - 1/5 + 1/5 - 1/6 +1/6 - 1/7 + 1/7 - 1/8
= 1/2 - 1/8
= 3/8
A=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
=1/2-1/3+1/4-1/5+1/5-1/6+1/7-1/8
=1/2-1/8
=4/8-1/8
=3/8
Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)
\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Giải:
1/12+1/20+1/30+...+1/9702
=1/3.4+1/4.5+1/5.6+...+1/98.99
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99
=1/3-1/99
=32/99
Chúc bạn học tốt!