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Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
=1-1/2+1/2-1/3+1/3-1/4+....+1/2014-1/2015
Trừ tất cả ta được 1-1/2015=2014/2015
=1-1/2+1/2-1/3+1/3-1/4+.....+1/2014-1/2015
=1-1/2015=2014/2015
\(A=\frac{1}{2}+\frac{1}{2.3}+..+\frac{1}{2017.2018}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=1-\frac{1}{2018}\)
\(A=\frac{2018}{2018}-\frac{1}{2018}\)
\(A=\frac{2017}{2018}\)
hok tốt!!
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
\(A=1-\frac{1}{2015}\)
\(A+\frac{1}{2015}=2x\Leftrightarrow1-\frac{1}{2015}+\frac{1}{2015}=2x\Leftrightarrow2x=1\Rightarrow x=\frac{1}{2}\)
A=1\(-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)
=> A= \(1-\frac{1}{2015}\)
A=\(\frac{2014}{2015}\)
A+\(\frac{1}{2015}=2x\)
<=>\(\frac{2014}{2015}+\frac{1}{2015}=2x\)
=>\(2x=1\)
\(=>x=\frac{1}{2}\)
\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)