Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
* Công thức : \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(A=\frac{3}{1.2.3}+\frac{3}{2.3.4}+...+\frac{3}{2015.2016.2017}\)
\(\Rightarrow A=3.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{2}-\frac{1}{4066272}\right)\)
\(\Rightarrow A=3.\left(\frac{2033136}{4066272}-\frac{1}{4066272}\right)\)
\(\Rightarrow A=3.\frac{2033135}{4066272}>3.\frac{1355424}{4066272}\)
\(\Rightarrow A>3.\frac{1}{3}\)
\(\Rightarrow A>1\)
Chúc bạn học tốt !!!
giong nhu dap an minh viet khi nay do
nho k cho minh voi nha
A=1/2(2/1.2.3+2/2.3.4+...+2/2014.2015.2016)~A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+...+1/2014.2015-1/2015.2016)~~A=1/2(1/1.2-1/2015.2016)~A=1/2(1/2-1/4062240)~A=1/2.2031119/4062240~A=203119/8124480. Dấu/= dấu gạch ps còn ~ là dấu xuống dòng. Còn bài này thì ko biết dung hay sai nua
2A=\(\frac{2}{1\cdot2\cdot3}\)+\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{2014\cdot2015\cdot2016}\)
2A=\(\frac{1}{1\cdot2}\)-\(\frac{1}{2\cdot3}\)+\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{2014\cdot2015}\)-\(\frac{1}{2015\cdot2016}\)
2A=\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)
A=(\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)):2
A=\(\frac{1}{2}\):2-\(\frac{1}{2015\cdot2016}\):2
A=\(\frac{1}{4}\)-\(\frac{1}{2015\cdot2016\cdot2}\)<\(\frac{1}{4}\)
Vậy A<\(\frac{1}{4}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)
\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2014.2015}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4058210}\right)\)
\(S=\frac{1}{2}.\left(\frac{2029105}{4058210}-\frac{1}{4058210}\right)\)
\(S=\frac{1}{2}.\frac{2029104}{4058210}\)
\(S=\frac{1014552}{4058210}\)
Chúc bạn học tốt !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
Ta có :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)
\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{2015.2016}\right):2\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{2015.2016}\)
\(\Rightarrow A< \frac{1}{4}\)
Vậy A < \(\frac{1}{4}\)
_Chúc bạn học tốt_
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2014+2015+2016}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{2014.2015.2016}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(2A=\frac{1}{1.2}-\frac{1}{2015.2016}\)
\(\Rightarrow2A< \frac{1}{1.2}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{4}\)
Vậy ....
= 1/2.(2/1.2.3+2/2.3.4+.....+2/50.51.52
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+....+1/50.51-1/51.52
=1/2.(1/1.2-1/51.52)
=1/2.(1/2-1/2652)
=1/2.1325/2652
=1325/5304
A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/50.51-1/51.52
A=1/1.2-1/51.52
phần còn lại tự giải nhé
Giải :
\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Study well ! >_<
A= 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + ....... + 1/2014 - 1/2015 - 1/2016
Rồi đoạn sau tự tính tiếp nhé :)) Đến đôạn này chắc trừ được
A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)
\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}-\frac{2}{2017}\)
\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2017}\)
\(\Leftrightarrow\)A=\(\frac{2016}{2017}\)
mk quên:Có \(\frac{2016}{2017}< \frac{1}{4}\) \(\Rightarrow\)S<\(\frac{1}{4}\)