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\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)
=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)
\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)
=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)
=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)
=>\(A>B\)
cách này mình tự nghĩ
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)
\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)
a/
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(A=2A-A=1-\frac{1}{2^{100}}< 1\)
b/
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)
Lời giải:
\(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
Dễ thấy $0< 2019^2< 2019^4\Rightarrow \frac{4}{2019^2}> \frac{4}{2019^4}$
$\Rightarrow A-B=\frac{4}{2019^2}-\frac{4}{2019^4}>0$
$\Rightarrow A>B$
thầy ơi vì sao \(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
Gọi biểu thức là A
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2019}}\)
⇒ 3A-A=2A=\(1+\frac{1}{3}+...+\frac{1}{3^{2019}}\)-\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2020}}\)
⇒ 2A=1-\(\frac{1}{3^{2020}}\)
⇒ A= \(\frac{1}{2}-\frac{1}{3^{2020}.2}\)
⇒ A< \(\frac{1}{2}\)