Ta có: \(P=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(3P=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}\)

=>3P+P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>4A=-1\(-\frac{1}{3^{2019}}\)

=>\(4A=\frac{-3^{2019}-1}{3^{2019}}\)

=>\(A=\frac{-3^{2019}-1}{4\cdot3^{2019}}\)

Ta có: \(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

\(=1+\frac{-3^{2019}-1}{4\cdot3^{2019}}-\frac{2020}{3^{2020}}=1+\frac{-3^{2020}-3-8080}{4\cdot3^{2020}}=1-\frac14-\frac{8083}{4\cdot3^{2020}}<\frac34\)

=>\(P<\frac{3}{16}\) (ĐPCM)

3^400<4^800

Nhớ giải thích

a: \(5^{300}=25^{150}\)

\(3^{450}=27^{150}\)

mà 25<27

nên \(5^{300}< 3^{450}\)

a: 

mà 25<27

nên 

mk sai đề một tí

A=2019 mũ 2020 + 1

trên 2019 mũ 2020 - 3

B=2019 mũ 2020 -1

trên 2019mũ 2020 - 5

   so sánh A và B

mn giúp mik vs

chưa hiểu đề hay là (-13)² hay -13² 

Ta có (-13)²=13^2=169

-13^2=-169

Vậy (-13)^2 > -13^2

bonu |-13^2|= | (-13)^2 |