3\(^4\)>\(^{4^3}\)

[100-99]\(^{2000}\)>[100+99]\(^0\)( vì theo dạng tổng quát ta có :a\(0\)=1 nên sẽ có điều như tớ làm nhé@@@@@@@@@)

A)Ta co:3^4=81

         4^3=64

        Vi 64<81

        =>3^4>4^3

B)Ta co:(100-99)^2000=1^2000=1

            (100+99)^0=199^0=1

            Vì:1=1

            =>(100-99)^2000=(100+99)^0

tk jup

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

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

Em nên gõ công thức trực quan để đề bài được rõ ràng nhé

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

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

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

gọi biểu thức trên là A , ta có :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

từ đó ta được :

\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)

help mik với 

chịuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

    A  = 2 +  2² + 2³ +...+2⁹⁹ + 2¹⁰⁰

   2A  =       2² + 2³ +...+ 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹

2A - A =      2¹⁰¹ - 2

   A      = 2¹⁰¹ - 2

  B        =  1 + 5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰

5B        =       5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰ + 5¹⁰¹

5B - B   =  5¹⁰¹ - 1

   4B     = 5¹⁰¹ - 1

     B      = \(\dfrac{5^{101}-1}{4}\)

     B     = 

lx

ko có câu hỏi bn ơi

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