Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{72}+\frac{1}{81}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}+\frac{1}{81}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(A=1-\frac{1}{9}+\frac{1}{81}=\frac{73}{81}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{81}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{81}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(=1-\frac{1}{9}+\frac{1}{81}\)
\(=\frac{8}{9}+\frac{1}{81}\)
\(=\frac{73}{81}\)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
a) = ( 5,4 - 4,4 ) + ( 6,5 - 5,5 ) + ( 7,6 - 6,6 ) + ( 8,7 - 7,7 )
= 1 + 1 + 1 + 1
= 4
b) = 9/10
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{4}{10}=\frac{2}{5}\)
Ủng hộ mk nha !!! ^_^
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+....+\frac{1}{9x10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
1/2 +1/(2.3) +1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7)
=(2-1)/2 + (3-2)/(2.3) +(4-3)/(3.4) + (5-4)/(4.5) + (6-5)/(5.6) + (7-6)/(6.7)
=1-1/2 +1/2-1/3+....+1/6-1/7
=1-1/7
=6/7
đặt biểu thức trên là A
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+....+\frac{7-6}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
1/6+1/12+....+1/90
= 1/2.3+1/3.4+1/4.5+.....+1/9.10
= 1/2-1/3+1/3-1/4+1/4-.......+1/9-1/10
= 1/2-1/10
=2/5
dấu chấm là nhân nha
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 +1/90
=1/2.3 + 1/3.4 +1/4.5 +1/5.6 +1/6.7 +1/7.8 +1/8.9
=1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9
=1/2 -1/9
=9/18 - 2/18
=7/18
misa
Đặt tên biểu thức là A ta có :
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+.....+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+......+0\)
\(=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}\)
\(=\dfrac{1}{3}\)
Sai đầu bài nhé, số cuối cùng phải là 110. Giải :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
= \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
=\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{11}\right)\)
=\(\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)
=\(\frac{9}{22}\)
Mình sửa đề 1 chút nha
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+........+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)