Vì : 

+) 68888/2422198 < 68888/275552 = 1/4

+) 46872/187488 = 1/4 < 46872/165564

=> 68888/2422198 < 46872/165564

\(\frac{68888}{2422198}\) < \(\frac{68888}{275552}\)=\(\frac{1}{4}\)=\(\frac{46872}{187488}\)<\(\frac{46872}{165564}\)nên\(\frac{68888}{2422198}\)<\(\frac{46872}{165564}\)

đó là cách 1

\(\frac{-2367}{3457214}\)và\(\frac{3457214}{2367}\)

Vì\(\frac{-2367}{3457214}< 0;\frac{3457214}{2367}>0\)

Nên\(\frac{-2367}{3457214}< \frac{3457214}{2367}\)

\(\frac{-2367}{3457214}\)và  \(\frac{3457214}{2367}\)

Vì \(\frac{-2367}{3457214}< 0;\frac{347214}{2367}>0\)

Nên  \(\frac{-2367}{3457214}< \frac{3457214}{2367}\)

                         Giải

Ở bài này, ta gọi số trung gian là 1.

Ta có \(\frac{-2367}{3457214}< 1\) và \(\frac{3457214}{2367}>1\)

nên \(\frac{-2367}{3457214}< 1< \frac{3457214}{2367}\)

Từ đó suy ra \(\frac{-2367}{3457214}< \frac{3457214}{2367}\)

giải:

Ta có:

\(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9-10}{10^{2011}}=\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}\)

\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9-10}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)

Vì \(\frac{10}{10^{2011}}< \frac{10}{10^{2010}}\rightarrow\frac{-10}{10^{2011}}>\frac{-10}{10^{2010}}\Rightarrow\frac{-9}{10^{2010}}+\frac{-9}{10^{2011}}+\frac{-10}{10^{2011}}>\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}+\frac{-10}{10^{2010}}\)

Vậy \(A>B\)( Bạn nhớ đọc kĩ lời giải nhé)

\(\frac{18}{91}\)\(\frac{23}{114}\)    MSC :10374

\(\frac{18\cdot114}{91\cdot114}=\frac{1052}{10374}\)

\(\frac{23\cdot91}{114\cdot91}=\frac{2093}{10374}\)

\(\frac{18}{91}< \frac{23}{114}\)

HOK TỐT .

18/91 < 18/90 = 1/5 = 23/115 < 23/114

=> 18/91 < 23/114

   Vậy 18/91 < 23/114

 \(\frac{18}{91}\)VÀ \(\frac{23}{114}\)=  \(\frac{18}{10374}\)VÀ  \(\frac{23}{10374}\)THẾ  \(\frac{18}{10374}\)<    \(\frac{23}{10374}\)

Hình như mẫu số hơi lớn, nếu sai thì thôi nha !

Ta có : \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

\(< \frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}\right)\)

\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(=\frac{1}{4}.\left(2-\frac{1}{n}\right)\)

\(=\frac{1}{2}-\frac{1}{4n}< 1\)

Vậy A < 1

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}.\)

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{4n^2}.\)

\(A=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}\right)\)

\(A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

So sánh \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....\)

\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n-1\right)}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n-1}+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(2-\frac{1}{n}\right)\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}\)

có \(\frac{1}{2}>\frac{1}{2}-\frac{1}{4n}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}< \frac{1}{2}\) mà \(\frac{1}{2}< 1\)

\(\Rightarrow A< 1\)

Ta có :

\(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{\left(2016^{2016}-1\right)+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)

\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{\left(2016^{2016}-3\right)+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)

Vì \(2016^{2016}-1>2016^{2016}-3\) nên \(\frac{3}{2016^{2016}-1}< \frac{3}{2016^{2016}-3}\)

\(\Rightarrow1+\frac{3}{2016^{2016}-1}< 1+\frac{3}{2016^{2016}-3}\)

\(\Rightarrow A< B\)

\(\frac{2323}{2424}=\frac{23.101}{24.101}=\frac{23}{24}\)

\(\frac{20132013}{20142014}=\frac{2013.10001}{2014.10001}=\frac{2013}{2014}\)

Ta có:

\(1-\frac{23}{24}=\frac{24}{24}-\frac{23}{24}=\frac{1}{24}\)

\(1-\frac{2013}{2014}=\frac{2014}{2014}-\frac{2013}{2014}=\frac{1}{2014}\)

Vì \(\frac{1}{24}>\frac{1}{2014}\) nên \(\frac{23}{24}< \frac{2013}{2014}\)

Vậy \(\frac{2323}{2424}< \frac{20132013}{20142014}\)

Tính phần bù với 1 nhé

mình cũng trình bày giống bạn "Muôn cảm súc"

thank you so much