Vì \(20^{10}-1>20^{10}-3\)

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

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

\(\Rightarrow B>A\)

\(\Rightarrow A< B\)

vậy A < B

1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)

      =(1-1/3)....0.....(1-9/5)

      =0

     =>đpcm.

b)ta xét:

1/2² = 1/2x2 < 1/1x2

.............

1/8² = 1/8x8 <1/7x8

=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8

<=> B <1 - 1/2 + 1/2  - 1/3  + ... + 1/7 - 1/8

<=> B < 1 - 1/8 = 7/8 < 1

=> B < 1 => đpcm

2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)

      Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)

Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)

=> A > B

   b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C

=> C > D

c)gọi 20¹⁰ là a

ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)

áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)

=> E > F

a

nAK.DNX. 0pwi9dOjkciopjopoijasd

1,10²⁰và 90¹⁰

ta có:+,10²⁰=(10²)¹⁰=100¹⁰

        +,90¹⁰=90¹⁰

vì 100¹⁰>90¹⁰=>10²⁰>90¹⁰

2,(1/16)¹⁰ và (1/2)⁵⁰

ta có:+, (1/16)¹⁰=(1/16)¹⁰

         +,(1/2)⁵⁰=(1/2⁵)¹⁰=(1/32)¹⁰

vì (1/16)¹⁰>(1/32)¹⁰=>(1/16)¹⁰>(1/2)⁵⁰

k mik nhé

\(a,\)  \(10^{20}=10^{10+10}=10^{10}.10^{10}\)

        \(90^{10}=9^{10}.10^{10}\)

  Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)

    \(\Rightarrow10^{20}>90^{10}\)

Vậy \(10^{20}>90^{10}\)

\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)

   \(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)

Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

                         ~~~~~~~~~~Hok tốt~~~~~~~~~~~

a) \(4^3\cdot32^4\)

\(=\left(2^2\right)^3\cdot\left(2^5\right)^4\)

\(=2^6\cdot2^{20}\)

\(=2^{26}\)

b) \(3^{20}\cdot9^{10}\cdot27^2\)

\(=3^{20}\cdot\left(3^2\right)^{10}\cdot\left(3^3\right)^2\)

\(=3^{20}\cdot3^{20}\cdot3^6\)

\(=3^{46}\)

c) \(3^{10}\cdot7^{10}\)

\(=\left(3\cdot7\right)^{10}\)

\(=21^{10}\)

d) \(6^{15}:6^{14}\)

\(=6^{15-14}\)

\(=6\)

e) \(28^3:7^3\)

\(=4^3\cdot7^3:7^3\)

\(=4^3\)

\(=2^6\)

\(\frac{10^{20}+1}{10^{22}+1}=\frac{10^{20}+\frac{1}{100}+\frac{99}{100}}{10^{22}+1}=\frac{1}{100}+\frac{99}{100\left(10^{22}+1\right)}\)

\(\frac{10^{22}+1}{10^{24}+1}=\frac{10^{22}+\frac{1}{100}+\frac{99}{100}}{10^{24}+1}=\frac{1}{100}+\frac{99}{100\left(10^{24}+1\right)}\)

Có \(10^{22}+1< 10^{24}+1\Rightarrow\frac{99}{100\left(10^{22}+1\right)}>\frac{99}{100\left(10^{24}+1\right)}\)

do đó \(\frac{10^{20}+1}{10^{22}+1}>\frac{10^{22}+1}{10^{24}+1}\).