\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009.2011}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{2009}-\frac{1}{2011}\)(Tối giản các phân số giống nhau)
\(2A=\frac{1}{1}-\frac{1}{2011}\)
\(2A=\frac{2010}{2011}\)
\(2A=\frac{2010}{2011}\Rightarrow A=\frac{2010}{2011}:2=\frac{2010}{4022}=\frac{1005}{2011}.\)
A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=2\cdot\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\left[\left(1-\frac{1}{2011}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)\right]\)
\(=\frac{1}{2}\cdot\left[\left(1-\frac{1}{2011}\right)+0+...+0\right]=\frac{1}{2}\cdot\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{2011}{2011}-\frac{1}{2011}\right)=\frac{1}{2}\cdot\frac{2010}{2011}=\frac{1\cdot2010}{2\cdot2011}=\frac{1005}{2011}\)
câu B cách làm cũng như thế, có điều là ví dụ như: 2=1*2; 6=2*3; 12=3*4
cách làm cũng tương tự, bạn tự suy nghĩ nha, chúc bạn học tốt!
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{2009.2011}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.........+\frac{1}{2009}-\frac{1}{2011}\)
\(2A=1-\frac{1}{2011}\)
\(A=\frac{2010}{2011}:2=\frac{1005}{2011}\)
Bài B Ta làm tương tự
Ax2 = 2/1.3 + 2/3.5 + 2/5.7 +......+ 2/2009.2011
Ax2 = 1-1/3 + 1/3-1/5 +1/5-1/7 +...+ 1/2009 - 1/2011
Ax2 = 1- 1/2011
Ax2=2010/2011
A=2010/2011 : 2
A= 1005/2011
