\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5\cdot10-\frac{2}{5}=5-\frac{2}{5}=\frac{23}{5}\)

\(\frac{\left(\sqrt{1\frac{1}{19}}-\sqrt{\frac{9}{16}}\right)}{9}=\frac{\sqrt{\frac{20}{9}}-\frac{3}{4}}{9}\).....

Đặt A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))

Ta có:A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))

A = \(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\).......\(\frac{99}{100}\)

A= \(\frac{1.3}{2.2}\). \(\frac{2.4}{3.3}\).\(\frac{3.5}{4.4}\).......\(\frac{9.11}{10.10}\)

A=\(\frac{1.2.3....9}{2.3.4....10}\).\(\frac{3.4.5....11}{2.3.4....10}\)

A= \(\frac{1}{10}\). \(\frac{11}{2}\)

A= \(\frac{11}{20}\)

Do 20> 19 => \(\frac{11}{20}\)< \(\frac{11}{19}\). Vậy A< \(\frac{11}{19}\)

Duyệt đi, chúc bạn học giỏi!

cách k ở đâu vậy

a) \(10^{30}=2^{30}.5^{30}=2^{30}.\left(5^3\right)^{10}=2^{30}.125^{10}\)

\(2^{100}=2^{30}.2^{70}=2^{30}.\left(2^7\right)^{10}=2^{30}.128^{10}\)

mà \(125^{10}< 128^{10}\)

\(\Rightarrow10^{30}< 2^{100}\)

b) \(5^{40}=\left(5^4\right)^{10}=625^{10}>620^{10}\)

\(5^{40}>620^{10}\)

c) \(8^{25}=\left(2^3\right)^{75}=2^{75}\)

\(16^{19}=\left(2^4\right)^{19}=2^{76}>2^{75}\)

\(\Rightarrow16^{19}>8^{25}\)

a,10³⁰ và 2¹⁰⁰

10³⁰=(10³)¹⁰=1000¹⁰

2¹⁰⁰=(2¹⁰)¹⁰=1024¹⁰

Vì 1000¹⁰<1024¹⁰ nên 10³⁰<2¹⁰⁰.

b,5⁴⁰ và 620¹⁰

5⁴⁰=(5⁴)¹⁰=625¹⁰>620¹⁰

=>5⁴⁰>620¹⁰.

c,8²⁵ và 16¹⁹

Nhân 8²⁵ và 16¹⁹ với 4 , ta được 

32²⁵ và 64¹⁹

32²⁵=(32⁵)⁵=33554432⁵

64¹⁹<64²⁰=(64⁴)⁵=16777216⁵

Vì 33554432⁵>16777216⁵ nên 8²⁵>16¹⁹

a 10^30 va 2^100

ta co 10^30 =(10^3)^10 =30^10

        2^100=(2^10)^10 =20^10

=>30^10>20^10 =>10^30>2^100

Bai nay de ma bn 

Bn tinh gia tri cac so ra xong so sanh!

Bn tu lam nha,mk chi goi y thui

nguyễn minh ngọc oi nếu thây dễ thì bạn thử lam hết xem có đúng ko ? rồi nói dễ

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

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

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

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

a) 16¹⁹ và 8²⁵ 

Ta có :

16¹⁹ = ( 2⁴ )¹⁹ = 2⁷⁶

8²⁵ = ( 2³ )²⁵ = 2⁷⁵

Vì 2⁷⁶ > 2⁷⁵ Nên 16¹⁹ > 8²⁵

b) 5³⁶ và 11²⁴

Ta có :

5³⁶ = ( 5³ )¹² = 125¹²

11²⁴ = ( 11² )¹² = 121¹²

Vì 125¹² > 121¹² Nên 5³⁶ > 11²⁴

1.

\(M=3^0+3^1+......+3^{50}.\)

\(\Rightarrow3M=3+3^2+.......+3^{51}\)

\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)

\(\Rightarrow2M=3^{51}-1\)

\(\Rightarrow M=\frac{3^{51}-1}{2}\)

2.

\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

                     \(8^{25}=\left(2^3\right)^5=2^{75}\)

Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)

\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

                      \(11^{24}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

\(N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(N>\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)

\(N>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{10}{22}>\frac{9}{22}\)

Vậy N > 9/22