\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)

\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)

Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)

Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).

Jdkdk

Jidkri

\(A=\dfrac{9}{8}-\dfrac{8}{9}+\dfrac{3}{24}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{19}{25}-\dfrac{1}{9}+\dfrac{2}{25}-\dfrac{1}{81}\)

\(=\dfrac{9}{8}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{1}{8}-\dfrac{8}{9}-\dfrac{1}{9}-\dfrac{1}{81}+\dfrac{19}{25}+\dfrac{2}{25}\)

\(=\dfrac{10}{8}+\dfrac{1}{4}-\dfrac{5}{16}-1-\dfrac{1}{81}+\dfrac{21}{25}\)

\(=\dfrac{20+4-5}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{19}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{32891}{16\cdot81\cdot25}\)

b: \(B=-\dfrac{1}{3}-\dfrac{8}{35}-\dfrac{2}{9}-\dfrac{1}{35}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{3}{7}\)

\(=\dfrac{-1}{3}-\dfrac{2}{9}-\dfrac{4}{9}-\dfrac{8}{35}-\dfrac{1}{35}+\dfrac{4}{5}+\dfrac{3}{7}\)

\(=\dfrac{-3-2-4}{9}+\dfrac{-9}{35}+\dfrac{28+15}{35}\)

\(=-1+\dfrac{-9+43}{35}=-1+\dfrac{34}{35}=-\dfrac{1}{35}\)

Ta có :

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)

ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)

             = 3/2^2.8/3^2 ... 99/10^2

             = 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2

             = 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10

             = 1/10 . 11/2 = 11/20 < 11/19

              Vậy M < 11/19

a) Ta có:

 8⁵¹> 8⁵⁰ (1)

8⁵⁰= 8^2.25=(8²)²⁵=64²⁵

Vì 48²⁵ < 64²⁵=> 48²⁵< 8⁵⁰  (2) 

Từ (1);(2) => 48²⁵< 8⁵¹

        b)   Ta có:

 5²⁰⁰⁰=5^2.1000 = (5²)¹⁰⁰⁰=25¹⁰⁰⁰

vì 25¹⁰⁰⁰> 10¹⁰⁰⁰=> 5²⁰⁰⁰> 10¹⁰⁰⁰

         c) 0,3¹⁰⁰ và 0,5²⁰¹

Ta có: 

0,5²⁰¹< 0,5²⁰⁰ (1)

0,5²⁰⁰=(0,5²)¹⁰⁰=(0,25)¹⁰⁰

Vì 0,3¹⁰⁰>0,25¹⁰⁰=>0,3¹⁰⁰> 0,5²⁰⁰ (2)

Từ (1) và (2) => 0,3¹⁰⁰> 0,5²⁰⁰

                 d)    32⁹ và 18¹³

Ta có:

32⁹=(2⁵)⁹=2⁴⁵

18¹³>16¹³=(2⁴)¹³=2^{52 (1)}

vì 2⁴⁵< 2⁵²=>  32⁹>16¹³ (2)

Từ (1);(2) => 32⁹> 18¹³