mk nhầm, là so sánh C và D nhé

Ta có \(7^{200}< 7^{205}\Rightarrow7^{200}+1< 7^{205}+1\Rightarrow\frac{7^{200}+1}{7^{202}+1}< \frac{7^{205}+1}{7^{202}+1}\)

vi 7²⁰⁰ + 1 < 7²⁰⁵ + 1 => \(\frac{7^{200}+1}{7^{202}+1}< \frac{7^{205}+1}{7^{202}+1}\)

                                  => \(A< B\)

^ là mũ nhé

2^300 = (2^3)^100=8^100 ; 3^200 = (3^2)^100 = 9^100

Mà 9 > 8 => 8^100 < 9^100

Vậy 2^300 < 3^200

99^20 = (99^2)^10 = 9801^10 và 9999^10

Mà 9999 > 9801 => 9801^10 < 9999^10

Vậy 99^20 < 9999^10

3^500 = (3^5)^100 = 243^100 

7^300 = (7^3)^100 = 343^100

Mà 343 > 243 => 343^100 > 243^100

Vậy 3^500 < 7^300

202^303 = (202^3)^101 = 8242408^101 ; 303^202 = (303^2)^101 = 91204

Vậy 202^303 > 303^202

a, A = 3⁵⁰⁰ = (3⁵)¹⁰⁰ = 243¹⁰⁰
B = 7³⁰⁰ = (7³)¹⁰⁰ = 343¹⁰⁰
Mà 243¹⁰⁰ < 343¹⁰⁰
=> A < B
@nguyễn thi trà giang

a) \(A=3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(B=7^{300}=\left(7^3\right)^{100}=343^{100}\)

Vì \(243^{100}< 343^{100}\Rightarrow3^{500}< 7^{300}\)

\(\Rightarrow A< B\)

b) \(A=303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(B=202^{303}=\left(202^3\right)^{101}=8242408^{101}\)

Vì \(91809^{101}< 8242408^{101}\Rightarrow303^{202}< 202^{303}\)

\(\Rightarrow A< B\)

c) \(A=3^{21}=3\cdot3^{20}=3\cdot\left(3^2\right)^{10}=3\cdot9^{10}\)

\(B=2^{31}=2\cdot2^{30}=2\cdot\left(2^3\right)^{10}=2\cdot8^{10}\)

Ta có: \(3>2;9^{10}>8^{10}\Rightarrow3\cdot9^{10}>2\cdot8^{10}\Rightarrow3^{21}>2^{31}\)

\(\Rightarrow A< B\)

em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

       D =           \(\dfrac{1}{7^2}\) - \(\dfrac{2}{7^3}\) + \(\dfrac{3}{7^4}\) - \(\dfrac{4}{7^5}\) +........+ \(\dfrac{201}{7^{202}}\) -  \(\dfrac{202}{7^{203}}\)

7 \(\times\) D  =  \(\dfrac{1}{7}\) -  \(\dfrac{2}{7^2}\) +  \(\dfrac{3}{7^3}\) - \(\dfrac{4}{7^4}\)  + \(\dfrac{5}{7^5}\) -.......- \(\dfrac{202}{7^{202}}\)

7D +D  =   \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)

         D = (  \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)) : 8

Đặt    B =      \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -........+\(\dfrac{1}{7^{201}}\).-\(\dfrac{1}{7^{202}}\) 

  7   \(\times\) B = 1 - \(\dfrac{1}{7}\)+\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) + \(\dfrac{1}{7^4}\) - \(\dfrac{1}{7^5}\) +.........- \(\dfrac{1}{7^{201}}\)

7B + B   =  1 - \(\dfrac{1}{7^{202}}\)

          B   =  ( 1 - \(\dfrac{1}{7^{202}}\)) : 8

         D  =  [ ( 1 - \(\dfrac{1}{7^{202}}\)): 8  - \(\dfrac{202}{7^{203}}\)] : 8 

          D = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)

a) Ta có:

+) \(\dfrac{1}{2}=\dfrac{3}{6}\)

+) \(\dfrac{1}{3}=\dfrac{2}{6}\)

+) \(\dfrac{2}{3}=\dfrac{4}{6}\)

=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)

hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)

b) Ta có:

+) \(\dfrac{4}{9}=\dfrac{56}{126}\)

+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)

+) \(\dfrac{3}{7}=\dfrac{54}{126}\)

=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)

hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)

c) Ta có:

+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)

+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)

=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)

hay \(\dfrac{27}{82}< \dfrac{26}{75}\)

d) Ta có:

+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)

+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)

=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)

hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)

\(\frac{201}{202}+\frac{202}{205}\)Và \(201+\frac{202}{202}+205\)

\(=\frac{201}{202}=\frac{201}{202}+\frac{1}{202}=\frac{202}{202}\)

\(\frac{202}{205}=\frac{202}{205}+\frac{3}{205}=\frac{205}{205}\)

\(201+1+205\)

Vậy \(1+1=2\)và \(407\)

=> \(\frac{201}{202}+\frac{202}{205}< 201+\frac{202}{202}+205\)

Ta có: \(\frac{201+202}{202+205}=\frac{201}{202+205}+\frac{202}{202+205}\)

Ta có: 202<202+205 => \(\frac{201}{202}>\frac{201}{202+205}\)(1)

205<202+205 => \(\frac{202}{205}>\frac{202}{202+205}\)(2)

Từ (1) và (2) => \(\frac{201}{202}+\frac{202}{205}>\frac{201+202}{202+205}\)