\(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}\) mà \(5^{200}=25^{100}\)

\(25^{100}< 27^{100}\Rightarrow3^{300}>5^{200}>5^{199}\)

Trong hai phân số cùng tử nếu mẫu nào lớn hớn thì phân số đó bé hơn.

Vậy : \(\frac{1}{5^{199}}>\frac{1}{3^{300}}\)

\(729.24^{100}=3^6.\left(2^3.3\right)^{100}=3^{106}.2^{300}\)

\(4^{300}=2^{300}.2^{300}\)

Ta có: \(2^{300}>2^{212}=\left(2^2\right)^{106}=4^{106}>3^{106}\)

\(\Rightarrow2^{300}.2^{300}>2^{300}.3^{106}\Rightarrow4^{300}>729.24^{100}\)

Vậy \(2^{300}+3^{300}+4^{300}>729.24^{100}\)

\(729 \cdot 24^{100} = 3^6 \cdot (3 \cdot 2^3)^{100} = 3^6 \cdot 3^{100} \cdot 2^{300} = 3^{106} \cdot 2^{300}\)

\(=3^6\cdot3^{100}\cdot2^{300}=(729\cdot3^{100})\cdot2^{300}\)

\(4^{300} = (2^2)^{300} = 2^{600} = 2^{300} \cdot 2^{300}\)

\(2^{300} = (2^3)^{100} = 8^{100}\)

\(\frac{8^{100}}{3^{100}} = \left(\frac{8}{3}\right)^{100}\)

mà 8/3>2,5=5/2

nên \(\left(\frac{8}{3}\right)^{100} > (2,5)^{100} = \left(\frac{5}{2}\right)^{100}\)

mà \((2,5)^{100}>2^{100}=(2^{10})^{10}=1024^{10}>1000^{10}=10^{30}>729\)

nên \(8^{100} > 729 \cdot 3^{100} \implies 2^{300} > 729 \cdot 3^{100}\)

=>\(2^{300} \cdot 2^{300} > (729 \cdot 3^{100}) \cdot 2^{300}\)

=>\(4^{300}>729\cdot24^{100}\quad(1)\)

Ta có: \(A = 2^{300} + 3^{300} + 4^{300} > 4^{300} \quad (2)\)

Từ (1),(2) suy ra \(2^{300} + 3^{300} + 4^{300} > 729 \cdot 24^{100}\)

Ta có:2³⁰⁰=2^2.100=(2²)¹⁰⁰=4¹⁰⁰

         3²⁰⁰=3^2.100=(3²)¹⁰⁰=9¹⁰⁰

          Vì:4¹⁰⁰ < 9¹⁰⁰

           =>2³⁰⁰ < 3²⁰⁰

ta có 2^300 = (2^3)^100 = 8^100

    3^200 = (3^2)^100 = 9^100 > 8 ^100

=> 2^300 < 3^200

vậy.........................

Ta có 199^4>200^3 => (199^4)^5>(200^3)^5 <=> 199^20>200^15

Ta có : 199²⁰ = ( 199⁴)⁵ = 1 568 239 201⁵

            200¹⁵ = ( 200³)⁵ = 8 000 000⁵

=> 199²⁰ < 200¹⁵

de ma bạn

199²⁰<200¹⁵