(100^99+99^100)^100

(100^100+99^100)^99

ta có : (100^99+99^100)^100=100^9900+99^10000

           (100^100+99^100)^99=100^9900+99^9900

=)100^9900=100^9900; 99^10000>99^9900(vì 10000>9900)

=)(100^99+99^100)^100>(100^100+99^100)^99

Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)

=>3S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{100}{3^{99}}\)

=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)

=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

=>\(4S=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>S<3/16

mà 3/16<3/15=1/5

nên S<1/5

2/5x10/7 x-3/4x10/7+5/2:2

2/5x10/7 x-3/4x10/7+5/2:2

\(2^{50}=\left(2^5\right)^{10}=32^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}\)

Suy ra: 2⁵⁰ > 5²⁰

b)

\(9^{200}=\left(9^2\right)^{100}=81^{100}\)

Suy ra: 99¹⁰⁰ > 81¹⁰⁰

\(5^{202}=\left(5^2\right)^{101}=25^{101}\)

\(2^{505}=\left(2^5\right)^{101}=32^{101}\)

Suy ra: 5²⁰² < 2⁵⁰⁵

a) 2^98 và 9^49

2^98 = (2^2)^49= 4^49

vì 4^49 < 9^49

nên 2^98 < 9^49

b) 3^44 và 4^33

3^44 = (3^4)^11= 81^11

4^33= ( 4^3)^11= 64^11

vì 81^11 > 64^11

nên 3^44 >4^33

Ta có : 2⁹⁸ = 2^2.49 = (2²)⁴⁹ = 4⁴⁹

Vì 4⁴⁹ < 9⁴⁹ => 2⁹⁸ < 9⁴⁹

Ta có : 3⁴⁴ = (3⁴)¹¹ = 81¹¹

            4³³ = (4³)¹¹ = 64¹¹

Vì 81¹¹ > 64¹¹ => 3⁴⁴ > 4³³

9 mũ 49=3⁹⁸>2⁹⁸

3⁴⁴=(3⁴)¹¹=81¹¹;4³³=(4³)¹¹=64¹¹=>3⁴⁴>4³³