2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

Trl:

Đây ko phải là bài lp 5 bn nhé.

Hok tốt!

trời đất toán lớp 5 khó bằng toán 6 lun á

đây ko phải lop5

a) Đặt A = 1 + 2 + 2² + ... + 2²⁰⁰⁸ (1)

=> 2A = 2 + 2² + 2³ + ... + 2²⁰⁰⁹ (2)

Lấy (2) trừ (1) theo vế ta có : 

2A - A = (2 + 2² + 2³ + ... + 2²⁰⁰⁹) - (1 + 2 + 2² + ... + 2²⁰⁰⁸)

       A = 2²⁰⁰⁹ - 1

Khi đó B = \(\frac{2^{2009}-1}{1-2^{2009}}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)

b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}\)

=> A - 1 = \(\frac{20^{10}+1-20^{10}+1}{20^{10}}=\frac{2}{20^{10}}\)

Lại có B = \(\frac{20^{10}-1}{20^{10}-3}\)

=> B - 1 = \(\frac{20^{10}-1-20^{10}+3}{20^{10}-3}=\frac{2}{2^{10}-3}\)

Vì \(\frac{2}{2^{10}}< \frac{2}{2^{10}-3}\)

=> A - 1 < B - 1

=> A < B

a) \(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)

Đặt \(Q=1+2+2^2+...+2^{2008}\)

\(2Q=2+2^2+2^3+...+2^{2009}\)

\(2Q-Q=2+2^2+2^3+...+2^{2009}-1-2-2^2-...-2^{2008}\)

\(\Rightarrow Q=2^{2009}-1\)

Ta thấy \(Q\) là số đối của \(2^{2009}-1\)

\(\Rightarrow B=-1\)

Vậy \(B=-1\).

b) Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

Ta lại có: \(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\) nên \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

\(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)

\(< =>2B=\frac{2+2^2+2^3+...+2^{2008}+2^{2009}}{1-2^{2009}}\)

\(< =>B=\frac{2^{2009}-1}{1-2^{2009}}=\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B

Mình làm câu a) nha!!!

+) \(A=2009^{2010}+2009^{2009}\)

        \(=2009^{2009}.\left(2009+1\right)\)

        \(=2009^{2009}.2010\)

+) \(B=2010^{2010}=2010^{2009}.2010\)

Vì \(2010^{2009}>2009^{2009}\)nên \(2010^{2009}.2010>2009^{2009}.2010\)hay \(B>A\)

Vậy \(A< B\)

Hok tốt nha^^

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì 10⁸-1 > 10⁸-3

=>\(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

=>\(1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\Rightarrow A< B\)

Ta có: A = 10⁸ + 2/ 10⁸ - 1 = 3/10⁸ - 1                           

              B = 10⁸ / 10⁸ - 3 = 3 / 10⁸ -3             

Vì 3 / 10⁸ - 1 < 3 / 10⁸ -3 nên           

Nên A< B

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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