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\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-....-\frac{1}{6.4}-\frac{1}{4.2}\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100.98}+\frac{1}{98.96}+....+\frac{1}{6.4}+\frac{1}{4.2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{98}+\frac{1}{98}-\frac{1}{96}+.....+\frac{1}{6}-\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\right)\)
\(\Rightarrow A=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)\Rightarrow A=\frac{1}{100}-\frac{1}{100}+\frac{1}{2}\Rightarrow A=\frac{1}{2}\)
\(A=\frac{1}{100}-\frac{1}{100.98}-\frac{1}{98.96}-...-\frac{1}{6.4}-\frac{1}{4.2}\)
\(A=\frac{1}{100}-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{96.98}+\frac{1}{98.100}\right)\)
\(A=\frac{1}{100}-\frac{1}{2.2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\left(1-\frac{1}{50}\right)\)
\(A=\frac{1}{100}-\frac{1}{4}.\frac{49}{50}\)
\(A=\frac{2}{200}-\frac{49}{200}=-\frac{47}{200}\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
a)Đặt \(L=\frac{1}{2^{2015}}+\frac{1}{2^{2014}}+...+\frac{1}{2^0}\)
\(2L=\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)
\(2L=2+1+...+\frac{1}{2^{2014}}\)
\(2L-L=\left(2+1+...+\frac{1}{2^{2014}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)
\(2L=2-\frac{1}{2^{2015}}\) thay vào ta có:
\(B=\frac{1}{2^{2016}}-\left(2-\frac{1}{2^{2015}}\right)=\frac{1}{2^{2016}}-2+\frac{1}{2^{2015}}\)
b)Ta có:\(\begin{cases}\left|x+1\right|\ge0\\\left|x+4\right|\ge0\end{cases}\)\(\Rightarrow\left|x+1\right|+\left|x+4\right|\ge0\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow3x\ge0\Rightarrow x\ge0\)
- Với \(x\ge0\) ta có
\(x+1+x+4=3x\)
\(\Rightarrow2x+5=3x\Rightarrow x=5\) (thỏa mãn)
Vậy x=5