a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

mà 17^19+1>17^18+1

nên A<B

b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)

\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)

2^2021-1<2^2022-1

=>1/2^2021-1>1/2^2022-1

=>-1/2^2021-1<-1/2^2022-1

=>C<D

a)          ta có công thức \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)

ta có \(N=\frac{5^2}{5.10}+\frac{5^2}{10.15}+...+\frac{5^2}{2005.2010}\)

\(N=5\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2005.2010}\right)\)

 \(N=5\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)(sử dụng quy tắc dấu ngoặc)

\(N=5\left[\frac{1}{5}-\left(\frac{1}{10}-\frac{1}{10}\right)-\left(\frac{1}{15}-\frac{1}{15}\right)-...-\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-0-0-...-0-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-\frac{1}{2010}\right]\)

\(N=5.\frac{401}{2010}\)

\(N=\frac{401}{402}\)

b)         \(M=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)

               ta thấy      \(\frac{1}{11}=\frac{1}{11}\)

                                \(\frac{1}{12}

kb đc 0

2 câu đầu tôi làm đc

a) Ta có :

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)

\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)

vậy A > 1

b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)

Vậy B > \(\frac{1}{2}\)

a/

\(27^{81}=\left(3^3\right)^{81}=3^{241}\)

\(81^{27}=\left(3^4\right)^{27}=3^{108}\)

\(\Rightarrow27^{81}=3^{241}>3^{108}=81^{27}\)

b/

\(5^{60}=\left(5^3\right)^{20}=125^{20}\)

\(7^{40}=\left(7^2\right)^{20}=49^{20}\)

\(\Rightarrow5^{60}=125^{20}>49^{20}=7^{40}\)

c/

\(11^{102}=\left(11^2\right)^{51}=121^{51}>121^{50}>99^{50}\)

gjup nhanh nha

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