đặt A=\(\frac{5^{12}+1}{5^{13}+1}\);B=\(\frac{5^{11}+1}{5^{12}+1}\);C= \(\frac{5^{11}-1}{5^{12}-1}\)

ta có:nhân A,B,C với 5 ta đc:\(5A=\frac{5\left(5^{12}+1\right)}{5^{13}+1}=\frac{5^{13}+5}{5^{13}+1}=\frac{5^{13}+1+4}{5^{13}+1}=\frac{5^{13}+1}{5^{13}+1}+\frac{4}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)

\(5B=\frac{5\left(5^{11}+1\right)}{5^{12}+1}=\frac{5^{12}+5}{5^{12}+1}=\frac{5^{12}+1+4}{5^{12}+1}=\frac{5^{12}+1}{5^{12}+1}+\frac{4}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)

\(5C=\frac{5\left(5^{11}-1\right)}{5^{12}-1}=\frac{5^{12}-5}{5^{12}-1}=\frac{5^{12}-1-4}{5^{12}-1}=\frac{5^{12}-1}{5^{12}-1}-\frac{4}{5^{12}-1}=1-\frac{4}{5^{12}-1}\)

vì 5¹³+1>5¹²+1>5¹²-1

=>\(\frac{4}{5^{12}-1}>\frac{4}{5^{12}+1}>\frac{4}{5^{13}+1}\)

\(\Rightarrow1+\frac{4}{5^{12}-1}>1+\frac{4}{5^{12}+1}>1+\frac{4}{5^{13}+1}\)

=>5C>5B>5A

=>C>B>A

bang nhau chu con gi nua

                    \(A=\frac{12}{5^{2012}}+\frac{18}{5^{2013}}\)

                   \(B=\frac{18}{5^{2012}}+\frac{12}{5^{2013}}\)

         =>  \(A=\frac{12}{5^{2012}}+\frac{12}{5^{2013}}+\frac{6}{5^{2013}}\)

               \(B=\frac{12}{5^{2012}}+\frac{12}{5^{2013}}+\frac{6}{5^{2012}}\)

             Mà \(\frac{6}{5^{2012}}>\frac{6}{5^{2013}}\)

                => \(B>A\)

                  Vậy B > A 

                Nhớ tk

\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)

Học good

\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)

\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)

\(=-\frac{1}{100}\cdot\frac{101}{2}\)

\(=-\frac{101}{200}\)

a)

i.Ta có: BCNN(12, 30) = 60

60 : 12 = 5; 60 : 30 = 2. Do đó:

\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)

ii.Ta có: BCNN(2, 5, 8) = 40

40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:

\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)

\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)

\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).

b)

i.Ta có: BCNN(6, 8) = 24

24 : 6 = 4; 24: 8 = 3. Do đó

\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)

ii. Ta có: BCNN(24, 30) = 120

120: 24 = 5; 120: 30 = 4. Do đó:

\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)