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Đặt \(A=\frac{10^{2018}+5}{10^{2018}-8};B=\frac{10^{2019}+5}{10^{2019}-8}\)
Ta có : \(A=\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)
\(B=\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)
Vì \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\)
\(\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\)
\(\Rightarrow A>B\)
Ta có:
10A=\(\frac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)
10B=\(\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1}{10^{2019}+1}+\frac{9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
do 1=1 và \(\frac{9}{10^{2018}+1}>\frac{9}{10^{2019}+1}\)
\(\Rightarrow\)A>B
Vậy A>B
chúc bạn học tốt!
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
a, Ta có : \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)
\(\Rightarrow\frac{13}{38}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)
b, Ta có: \(3^{301}>3^{300}=\left(3^3\right)^{100}=27^{100}\left(1\right)\)
\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\left(2\right)\)
Do \(25^{100}< 27^{100}\Rightarrow5^{200}< 3^{300}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow5^{199}< 5^{200}< 3^{300}< 3^{301}\Rightarrow5^{199}< 3^{301}\)
c, Ta có: \(\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)
\(\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)
Do \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)