Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)

Vậy A < B

lolllllo

Vì \(\frac{10^{18}+1}{10^{19}+1}< 1\Rightarrow B=\frac{10^{18}+1}{10^{19}+1}< \frac{10^{18}+1+9}{10^{19}+1+9}\)

\(\Rightarrow B< \frac{10^{18}+10}{10^{19}+10}\)

\(\Rightarrow B< \frac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)

\(\Rightarrow B< \frac{10^{17}+1}{10^{18}+1}\)

\(\Rightarrow B< A\)

Vậy A > B.

So sánh 2 phân số sau  $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10 

kick dzô chữ xanh là được!! OK

Ta có : 

10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)

         = \(\frac{10^{2012}+10}{10^{2012}+1}\)

         = \(\frac{10^{2012}+1+9}{10^{2012}+1}\)

         = \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)

         = 1 - \(\frac{9}{10^{2012}+1}\)

10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)

          = \(\frac{10^{2013}+10}{10^{2013}+1}\)

          = \(\frac{10^{2013}+1+9}{10^{2013}+1}\)

          = 1 - \(\frac{9}{10^{2013}+1}\)

Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\)  nên 10.A > 10.B

=> A >B 

Vậy ...........

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

               \(=\frac{-9}{10^{2010}}-\frac{9}{10^{1011}}-\frac{1}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}\)

Ta thấy : \(\frac{10}{10^{2010}}< \frac{19}{10^{2010}}\Rightarrow\frac{-10}{10^{2010}}>\frac{-19}{10^{2010}}\)

            \(\Rightarrow\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)

Hay \(A>B\)

Vậy ...

Ta có:

\(\dfrac{1}{20^2}< \dfrac{1}{20\cdot19}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\dfrac{1}{21^2}< \dfrac{1}{20\cdot21}=\dfrac{1}{20}-\dfrac{1}{21}\)

\(...\)

\(\dfrac{1}{30^2}< \dfrac{1}{29\cdot30}=\dfrac{1}{29}-\dfrac{1}{30}\)

\(\Rightarrow A< \dfrac{1}{19}-\dfrac{1}{30}< \dfrac{1}{19}\)

\(10A=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)

\(10B=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)

Vì \(\frac{9}{10^{2015}+1}>\frac{9}{10^{2017}+1}\Rightarrow10A>10B\Rightarrow A>B\)

a=1.1 b=1.1 a=b

\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)

vì 10¹²-1>10¹¹+1

=>\(\frac{9}{10^{12}-1}<\frac{9}{10^{11}+1}\)

=>A<B

Ta có:\(10A=\frac{10\left(10^{11}-1\right)}{10^{12}-1}=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)

\(10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)

Vì \(1-\frac{9}{10^{12}-1}<1+\frac{9}{10^{11}+1}\)

Nên A<B

a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A=2+2^2+2^3+2^4+...+2^{201}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A-A=2^{101}-1\)

\(A=2^{201}-1\)

Ta có 2²⁰¹ > 2²⁰¹ - 1 => B > A => 2²⁰¹ > 1 + 2 + 2² + 2³ +...+ 1¹⁰⁰

b) 2¹⁰⁰ = 2³¹ . 2⁶³ . 2⁶ = 2³¹ . (2⁹)⁷ . (2²)³ = 2³¹ . 512⁷ . 4³ (1)

10³¹ = 2³¹ . 5²⁸ . 5³ = 2³¹ . (5⁴)⁷ . 5³ = 2³¹ . 625⁷ . 5³ (2)

Từ (1) , (2) => 2³¹ . 512⁷ . 4³ < 2³¹ . 625⁷ . 5³ ( vì 512⁷ < 625⁷ và 4³ < 5³ )

=> 2¹⁰⁰ < 10³¹

e) Ta có:

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰

10³⁰ = (10³)¹⁰ = 1000¹⁰
Vì 1024¹⁰ > 1000¹⁰ => 2¹⁰⁰ > 10³⁰