2²⁵⁰ > 3¹⁰⁰

Ta có :

`2^250 = ( 2^2 )^{125} = 4^{125}`

Do `3^{100} < 4^{100}<4^{125} => 3^{100}<4^{125}=>2^{250}>3^{100}`

Vậy `2^{250}>3^{100}`

Xét 2⁵>2⁴=16 mà 16>9=3²

⇒ 2⁵>3²

⇒ (2⁵)⁵⁰>(3²)⁵⁰

⇒ 2²⁵⁰>3¹⁰⁰

Vậy 2²⁵⁰>3¹⁰⁰

\(4^{250}=\left(2^2\right)^{250}=2^{500}=\left(2^5\right)^{100}=32^{100}\)

Vì \(32^{100}>25^{100}\)nên \(4^{250}>25^{100}\)

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

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

\(=-\frac{\left(2^2-1\right)\times\left(3^2-1\right)\times...\times\left(100^2-1\right)}{2^2\times3^2\times...\times100^2}\)

\(=-\frac{\left(1\times3\right)\times\left(2\times4\right)\times...\times\left(99\times101\right)}{2^2\times3^2\times...\times100^2}\)

\(=-\frac{\left(1\times2\times...\times99\right)\times\left(3\times4\times...\times101\right)}{\left(2\times3\times...\times100\right)\times\left(2\times3\times...\times100\right)}\)

\(=-\frac{1\times101}{100\times2}=-\frac{101}{200}< -\frac{1}{2}\)

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)

\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)

\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

Vì 20 < 21 nên 11/20 > 11/21

Vậy ..... 

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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\)

\(A=\frac{1}{20}\)

Vì 20 < 21 nên 1/20 > 1/21 

Vậy ............

de nhu an chuoi

Áp dụng a /b > 1 => a/b > a+m/b+m (a;b;m thuộc N*)

Ta có:

\(\frac{100^{10}-1}{100^{10}-3}>\frac{100^{100}-1+2}{100^{10}-3+2}\)

                    \(>\frac{100^{100}+1}{100^{10}-1}\)

Áp dụng a/b > 1 => a/b > a + m/b ( a; b; m thuộc N*

ban co the so sanh bang cach quy dong mau hoac tu vi cac cach khac ko giai dc.

Ta có A = 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰

=> 2A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹

Khi đó 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹) - (1 + 2 + 2² + 2³ + ... + 2¹⁰⁰)

             => A  = 2¹⁰¹ - 1 

Vì 2¹⁰¹ - 1 < 2¹⁰¹

=> A < B

Vậy A < B

A = 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰

=> 2A = 2( 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰ )

           = 2 + 2² + 2³ + ... + 2¹⁰¹

=> A = 2A - A

         = 2 + 2² + 2³ + ... + 2¹⁰¹ - ( 1 + 2 + 2² + 2³ + ... + 2¹⁰⁰ )

         = 2 + 2² + 2³ + ... + 2¹⁰¹ - 1 - 2 - 2² - 2³ - ... - 2¹⁰⁰ 

         = 2¹⁰¹ - 1 < 2¹⁰¹

=> A < B