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a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt
bn ơi mk nghĩ bn nên tôn trọng mk một chút! Nếu bn giúp đc thì mk cảm ơn rất nhiều. Còn bn không làm đc thì để cho người khác làm! bn ko thích làm thì mk cx ko mong bn giải nửa chừng như vậy, mk vừa ko hiểu j mà còn bị tự ái khi bn nói như vậy, mong bn hiểu!!mk góp ý thật lòng, ko chỉ đối với mk mà với những bn khác cx zậy!!
a) Ta có:
3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)
A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)
Lấy (1) - (2) ta được:
1-\(\dfrac{1}{3^{100}}\)
b) Ta xét:
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)
Ta có:
2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)
=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)
Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)
=\(\dfrac{370}{741}\)
Nếu bn cảm thấy mk đúng tick cho mk nhé!
hôm qua cô giảng cho mình bài này không cần tính đâu
Gọi tổng là A
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{17.18.19}\)
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{17.18.19}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{17.18}-\dfrac{1}{18.19}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{18.19}\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{18.19}\right)\)
A=\(\dfrac{1}{2}.\dfrac{18.19-2}{2.18.19}\) < \(\dfrac{1}{4}\)
A=\(\dfrac{18.19-2}{2.2.18.19}\) < \(\dfrac{18.19}{2.2.18.19}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)
2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))
2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)
A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)
Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)
Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )
Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
\(linh_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}.\dfrac{9}{20}=\dfrac{9}{40}\)
\(linh_2=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\)\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{1}{2}.\dfrac{22}{45}=\dfrac{11}{45}\)
a/ \(G=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{2}-\dfrac{1}{20}\)
\(\Leftrightarrow2G=\dfrac{9}{20}\)
\(\Leftrightarrow G=\dfrac{9}{40}\)
b/ \(H=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+.....+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow2H=\dfrac{22}{45}\)
\(\Leftrightarrow H=\dfrac{22}{90}\)
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
Ta có :
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)
...
Do đó :
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
Vậy :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{20}\)
\(A=\dfrac{20}{20}-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
A = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{18.19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{18.19}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)
A = \(\dfrac{189}{380}\)
(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha
)