10A=10*\(\frac{10^{2006}+1}{10^{2007}+1}\)                             10B=10*\(\frac{10^{2007}+1}{10^{2008}+1}\)                           

10A=\(\frac{10^{2007}+1+9}{10^{2007}+1}\)                                10B=\(\frac{10^{2008}+1+9}{10^{2008}+1}\)

10A=1+\(\frac{9}{10^{2007}+1}\)                                10B=1+\(\frac{9}{10^{2008}+1}\)

Vì \(\frac{9}{10^{2007}+1}\)>\(\frac{9}{10^{2008}+1}\)=>1+\(\frac{9}{10^{2007}+1}\)>1+\(\frac{9}{10^{2008}+1}\)

Nên 10A>10B=>A>B

Ta có: \(A=\frac{10^{2006}+1}{10^{2007}+1}\)

\(=>10A=\frac{10^{2007}+10}{10^{2007}+1}=\frac{10^{2007}+1+9}{10^{2007}+1}=\frac{10^{2007}+1}{10^{2007}+1}+\frac{9}{10^{2007}+1}=1+\frac{9}{10^{2007}+1}\)

            \(B=\frac{10^{2007}+1}{10^{2008}+1}\)

\(=>10B=\frac{10^{2008}+10}{10^{2008}+1}=\frac{10^{2008}+1+9}{10^{2008}+1}=\frac{10^{2008}+1}{10^{2008}+1}+\frac{9}{10^{2008}+1}=1+\frac{9}{10^{2008}+1}\)

Vì \(10^{2007}+1< 10^{2008}+1=>\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}=>1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}=>10A>10B=>A>B\)

Cho B = \(\frac{10^{2007}+1}{10^{2008}+1}\)

Rõ ràng B < 1 nên theo B, nếu \(\frac{a}{b}< 1\) thì \(\frac{a+n}{b+n}>\frac{a}{b}\) => B < \(\frac{\left(10^{2007}+1\right)+9}{\left(10^{2008}+1\right)+9}=\frac{10^{2007}+10}{10^{2008}+10}\)

Do đó B < \(\frac{10^{2007}+10}{10^{2008}+10}=\frac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\frac{10^{2006}+1}{10^{2007}+1}\)

=> A > B

tick đi mình giải cho,dễ ẹc à.

Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)

\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)

hay \(A=3-\dfrac{2}{3^9+1}\)

Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)

\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)

hay \(B=3-\dfrac{2}{3^8+1}\)

Ta có: \(3^9+1>3^8+1\)

\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)

hay A>B

Câu này á ???

\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)

\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

=>B<1

=>A>B

\(\dfrac{11}{10}< \dfrac{9}{30}\)

\(\dfrac{6}{7}>\dfrac{3}{5}\)

\(\dfrac{25}{100}< \dfrac{3}{4}\)

a \(\dfrac{11}{10}>\dfrac{3}{10}\)

b \(\dfrac{30}{35}>\dfrac{21}{35}\)

c \(\dfrac{1}{4}< \dfrac{3}{4}\)

a, \(\dfrac{11}{10}< \dfrac{9}{30}\)

b, \(\dfrac{6}{7}>\dfrac{3}{5}\)

c, \(\dfrac{25}{100}< \dfrac{3}{4}\)

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K mk nhé

a)\(3\frac{4}{10}>2\frac{9}{10}\)

b)\(3\frac{4}{10}< 3\frac{9}{10}\)

c)\(5\frac{1}{10}>2\frac{9}{10}\)

d)\(3\frac{4}{10}=3\frac{2}{5}\)

Hok tốt!~

a) 3 9/10 > 2 9/10

b) 3 4/10 < 3 9/10

c) 5 1/10 > 2 9/10

d) 3 4/10 = 3 2/5