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

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

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 \(2^{300} + 3^{300} + 4^{300} > 729 \cdot 24^{100}\)

\(3^{300}+4^{300}\)

\(=27^{100}.64^{100}\)

\(=1728^{100}>3.24^{100}\)

Lớn hơn nhé !

Ta có :

\(3.24^{100}=3.3^{100}.8^{100}=3^{101}.8^{100}\)

Xét : \(4^{300}\)và \(3^{101}.8^{100}\)ta có : 

\(4^{300}=2^{300}.2^{300}=\left(2^2\right)^{150}.\left(2^3\right)^{100}=\)\(4^{150}.8^{100}\)

Vì \(8^{100}=8^{100}\)và \(4^{150}>3^{101}\Rightarrow4^{300}>3^{101}.8^{100}\)

\(\Rightarrow4^{300}+3^{400}>3.24^{100}\)

Dấu <

Ta có: \(3.24^{100}=3.3^{100}.8^{100}=3^{101}.8^{100}\)

Xét \(4^{300}\)và \(3^{101}.8^{100}\)có:\(4^{300}=2^{300}.2^{300}=\left(2^2\right)^{150}.\left(2^3\right)^{100}=4^{150}.8^{100}\)

Vì 8¹⁰⁰=8¹⁰⁰ và 4¹⁵⁰>3¹⁰¹ nên 4³⁰⁰>3¹⁰¹.8¹⁰⁰

nên 3.24¹⁰⁰<4³⁰⁰+3⁴⁰⁰