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\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Áp dụng công thức : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
S=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
S=1-\(\frac{1}{2010}\)
S=\(\frac{2009}{2010}\)
k nha bn
\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}+\frac{1}{2009\times2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
Vậy \(S=\frac{2009}{2010}\)
Học tốt #
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1/1-1/100
=100/100-1/100
=99/100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{99}{100}\)
~~~
#Sunrise
\(=1\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=\(1-\frac{1}{9}\)
=\(\frac{8}{9}\)
OK XONG NHỚ CHO MIK NHA
\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}\)
=\(\frac{8}{9}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\left(1-\frac{1}{10}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)\)
\(=\left(\frac{10}{10}-\frac{1}{10}\right)+0+...+0=\frac{9}{10}\)
...
= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
= 1/1-1/10
= 9/10
Bây giờ mình cho tổng là A đi
tìm quy luật \(\frac{1}{1x2}=\frac{2-1}{1x2}=\frac{1}{1}-\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{2x3}=\frac{3-2}{2x3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3x4}=\frac{1}{3}-\frac{1}{4}\)
tương tự các số sau cũng như vậy
Giải
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
trừ \(\frac{1}{2}\)cộng \(\frac{1}{2}\)hết
chỉ còn lại số hạng đầu tiên là 1 và số hạng cuối cùng là\(\frac{1}{10}\)
vậy A = 1 -\(\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
vậy Đ/S : \(\frac{9}{10}\)
Ta có: 1- 1
9/10
de thoi ma ban chi can lay phan so dau la 1/1-1/10
Đáp số: \(\frac{9}{10}\).
Đúng 100% luôn!
Ai tk cho mình mình tk lại.
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
kết quả là 9/10
Cho S = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/8x9 + 1/9x10
S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9 + 1/9 - 1/10
S = 1 - 1/10
S = 9/10
=1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .......+ 1/8 - 1/9 + 1/9 -1/10
=1 - 1/10
=9/10
\(=\frac{9}{10}\)nhé
ta có xông thức : \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n\times\left(n+1\right)}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdot\cdot\cdot+\frac{1}{8\times9}+\frac{1}{9\times10}=\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
vì \(\left(-\frac{1}{2}+\frac{1}{2}\right),\left(-\frac{1}{3}+\frac{1}{3}\right),\cdot\cdot\cdot,\left(-\frac{1}{9}+\frac{1}{9}\right)\) nên ta sẽ lượt bỏ có số như vậy thì còn :
\(=1-\frac{1}{10}=\frac{9}{10}\)
Tổng của dãy : \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdot\cdot\cdot+\frac{1}{8\times9}+\frac{1}{9\times10}\)=\(\frac{1}{10}\)
= 1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9+1/9-1/10
=1/1-1/10
=9/10
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)