a/

2020.2021=(2019+1)(2022-1)=

=2019.2022-2019+2022-1=2019.2022+2>2019.2022

b/

\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)

c/

\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)

\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)

\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)

d/

\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)

\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)

\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)

có mà nguyễn thị thối

cái nào mũ lớn hơn thì nó lớn lơn

a)Ta có : \(3^{39}=\left(3^{13}\right)^3=1594323^3\)
  \(11^{21}=\left(11^7\right)^3=19487171^3\)
Vì \(1594323< 19487171\)
\(=>1594323^3< 19487171^3\)
\(=>3^{39}< 11^{21}\)
Vậy \(3^{39}< 11^{21}\)
   b)Ta có : \(199^{20}=\left(199^4\right)^5=1568239201^5\)
                \(2002^{15}=\left(2002^3\right)^5=8024024008^5\)
Vì \(1568239201< 8024024008\)
\(=>1568239201^5< 8024024008^5\)
\(=>199^{20}< 2002^{15}\)
Vậy \(199^{20}< 2002^{15}\)
c) Ta có:\(125^{90}=\left(125^3\right)^{30}=1953125^{30}\)
            \(25^{120}=\left(25^4\right)^{30}=390625^{30}\)
Vì \(1953125>390625\)
\(=>1953125^{30}>390625^{30}\)
\(=>125^{90}>25^{120}\)
Vậy \(125^{90}>25^{120}\)
d)Ta có : \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
              \(7^{300}=\left(7^3\right)^{100}=343^{100}\)
Vì \(243< 343\)
\(=>243^{100}< 343^{100}\)
\(=>3^{500}< 7^{300}\)
Vậy \(3^{500}< 7^{300}\)
Phù , cuối cùng cũng viết xong . Mỏi tay quá ! À , chúc bạn học tốt nhé !

\(\)

Có 5¹⁴⁹=5⁹⁹⁺⁵⁰=5⁹⁹ x 5⁵⁰

mà 5⁹⁹ > 11⁹⁹

=> 5¹⁴⁹ > 11⁹⁹

k hộ mik nha!

Hình như bạn sai. 5^99<11^99 mà.

Mình có cách này, không thuận tiện lắm nhưng có thể nói là tạm dùng được để so sánh.

Ta có:

\(11^{99}\)

\(=\left[11^{99}\div10^{99}\div\left(\frac{11}{10}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left[\left(\frac{11}{10}\right)^{99}\div\left(\frac{11}{10}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left(\frac{11}{10}\right)^{50}\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(5^{149}\)

\(=\left[5^{149}\div10^{99}\div\left(\frac{11}{10}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left[5^{149}\div5^{99}\div2^{99}\div\left(\frac{11}{10}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left[5^{50}\div2^{50}\div2^{49}\div\left(\frac{11}{10}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left\{\left(5^{50}\div2^{50}\right)\div\left[2^{49}\cdot\left(\frac{11}{10}\right)^{49}\right]\right\}\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left[\left(\frac{5}{2}\right)^{50}\div\left(\frac{11}{5}\right)^{49}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(=\left[\left(\frac{5}{2}\right)^{50}\div\left(\frac{11}{5}\right)^{50}\cdot\frac{11}{5}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

\(\left[\left(\frac{25}{22}\right)^{50}\cdot\frac{11}{5}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\)

Mà \(\frac{25}{22}>\frac{11}{10}\Rightarrow\left[\left(\frac{25}{22}\right)^{50}\cdot\frac{11}{5}\right]\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]>\left(\frac{11}{10}\right)^{50}\cdot\left[10^{99}\cdot\left(\frac{11}{10}\right)^{49}\right]\Rightarrow5^{149}>11^{99}\)

Khi nào nghĩ được cách hay hơn mình sẽ đăng tiếp. k mình nha.

1, A = 2⁹¹ = 2^7.13 = (2¹³)⁷ = 8192⁷

B = 5³⁵ = 5^5.7 = (5⁵)⁷ = 3125⁷

Vì 3125 < 8192

=> 3125⁷ < 8192⁷

=> B < A

2.Ta có:

 A=11+11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰.

11A=11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰+11²⁰¹.

11A-A=11²⁰¹-11.

10A=11²⁰¹-11.

A=(11²⁰¹-11):10

Quan sát 2 vế A và B thì ta thấy rõ ràng vế A<B hay B>A.

\(a,\frac{12345}{12342}=1+\frac{3}{12342}\)

    \(\frac{23457}{23455}=1+\frac{2}{23455}\)

Vì \(\frac{3}{12342}>\frac{2}{23455}\Rightarrow\frac{12345}{12342}>\frac{23457}{23455}\)

\(b,\frac{149}{150}=1-\frac{1}{150}\)

        \(\frac{4}{11}=1-\frac{7}{11}\)

Vì \(\frac{1}{150}< \frac{7}{11}\Rightarrow\frac{149}{150}>\frac{4}{11}\)

12345/12342 = 1 + 3/12342 = 1 + 1/ 4114 = 1 + 2/8228

23457/23455 = 1 + 2/23455

Có 1 + 2/8228 > 1 + 2/23455 => 12345/12342 > 23457/23455.

149/450 < 150/450 = 1/3

4/11 > 4/12 = 1/3

Vậy 149/450 < 4/11