Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\dfrac{1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{11}{12}\) + \(\dfrac{19}{20}\) + \(\dfrac{29}{30}\) + \(\dfrac{41}{42}\) + \(\dfrac{55}{56}\)
A = (1 - \(\dfrac{1}{2}\)) + ( 1 - \(\dfrac{1}{6}\)) + (1 - \(\dfrac{1}{12}\)) + (1 - \(\dfrac{1}{20}\)) +(1-\(\dfrac{1}{30}\))+(1-\(\dfrac{1}{42}\))+(1-\(\dfrac{1}{56}\))
A = (1 + 1+1 + 1 + 1+1+1)- (\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\))
A = 7 - (\(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{8}\))
A = 7 - \(\dfrac{7}{8}\)
A = \(\dfrac{49}{8}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
A=1/2+ 5/6 + 11/12 + 19/20 + 29 30 + 41/42 + 55/56 + 71/72 + 89/90
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72
=10-(1/2+1/6+..+1/110)
=10-(1/1x2+1/2x3+...+1/10x11)
=10-(1-1/2+1/2-1/3+...+1/10-1/11)
=10-(1-1/11)
=10-10/11
=100/11
\(A=\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
Gọi \(A=9-B\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(A=9-\frac{9}{10}\)
\(A=\frac{90-9}{10}=\frac{81}{10}\)
Ko đúng hơi tiếc :D
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
= 1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
= 9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
= 9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
= 9 – ( 1-1/2+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+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10 = 81/10
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110
= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 .....+1 - 1/110
= 10 - ( 1/2 + 1/6 + ...+ 1/110)
= 10 - ( 1 - 1/ 2+ 1/2 - 1/ 3+ 1/3 - 1/4 ....+ 1/10 - 1/11)
= 10 - (1 - 1/11)= 10 - 10/11
= 100/11
Câu hỏi của Try Build Gundam - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=3A-A\)
\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(2A=\frac{3}{4}-\frac{1}{927}\)
\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)
\(A=\frac{182}{243}:\frac{1}{2}\)
\(A=\frac{364}{243}\)
A=1/4+1/12+1/36+1/108+1/324+1/972
=243/972+81/972+27/972+9/972+3/972+1/972
=364/972
=91/243