\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\);\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)
Chứng minh rằng \(\frac{A}{B}\)thuộc Z
\(A=\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
Tìm A
Bó tay, sai đề rồi bn à, nếu tính đc thì cũng dài dòng lắm...........
Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}+\frac{1}{1005.2006}+\frac{1}{1006.2006}+...+\frac{1}{2006.2006}\)
Tính A chia B
cho A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2005.2006}\)và B=\(\frac{1}{1008}\)+\(\frac{1}{1009}\)+...+\(\frac{1}{2016}\). Tính B-A
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{2005}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\)\(-\frac{1}{1}-\frac{1}{2}-...-\frac{1}{1003}\)
= \(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2005}+\frac{1}{2006}\)
(=) B - A = \(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}+\frac{1}{2016}\)- \(\frac{1}{1004}-\frac{1}{1005}-...-\frac{1}{2005}-\frac{1}{2006}\)
= \(\frac{1}{2007}+\frac{1}{2008}+...+\frac{1}{2016}-\) \(\frac{1}{1004}-\frac{1}{1005}-\frac{1}{1006}-\frac{1}{1007}\)
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+............+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+.....+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)
\(=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)(1)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+....+\frac{1}{2006.1004}\)
\(\Rightarrow\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
\(=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)(2)
Thế (1) và (2) vào ta có:
\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}\)
Cho:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.....+\frac{1}{2006.1004}\)
Hãy tính \(\frac{A}{B}\)
1, Cho
A = \(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)
B = \(\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.......+\frac{1}{2006.1004}\)
Hãy Tính \(\frac{A}{B}\)
A=1−12+13−14+...+12005−12006=(1+12+...+12006)−(1+12+..+11003)=11004+11005+...+12006" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:41.489em; padding:1px 0px; position:relative; white-space:nowrap; width:749.281px; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
13010.B=11004+12006+11005+12005+...+11004=11505(11004+11005+...+12006)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
Suy ra A/B = 1505
Tham khảo nha
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2005.2006}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2005}-\frac{1}{2006}\)
\(=1-\frac{1}{2006}\)
\(=\frac{2005}{2006}\)
Cho
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
Tìm \(\frac{A}{B}\)
Chắc chắn 100% đề bài đúng!!
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
=>3010B=\(\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\cdot\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
=>B=\(\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)
=>\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(B=\frac{a^{10}+b^{20}+c^{30}}{abc}\) với \(a=1990;b=2016\)
\(C=1^2-2^2+3^2-4^2+...+2015^2-2016^2\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
=> \(A=\frac{1}{1}-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}\)
\(A=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
Vậy \(A=\frac{2005}{2006}\)
1;Tính:
A=22+42+62+82+...+1002.
B=13+23+33+43+...+1003.
2;Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.2006}\)
Tính A : B
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