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Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)
\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)
\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)
Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)
Vì \(\frac{-3}{2}< 0\)nên A < B
Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)
\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)
\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)
\(=-2\)
\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)
\(\Rightarrow B=0\)
Vì -2 < 0 nên A < B
Vậy A < B
\(2^{300}+3^{300}+4^{300}-729.24^{100}=\)
\(=2^{300}+3^{300}+\left(2^2\right)^{300}-3^6.\left(2^3.3\right)^{100}=\)
\(=2^{300}+3^{300}+2^{600}-2^{300}.3^{106}=\)
\(=2^{300}\left(1+2^{300}-3^{106}\right)+3^{300}\)
Ta có
\(2^{300}=\left(2^2\right)^{150}=4^{150}>3^{150}>3^{106}\Rightarrow2^{300}-3^{106}>0\)
\(\Rightarrow2^{300}\left(1+2^{300}-3^{106}\right)+3^{300}>0\)
\(\Rightarrow2^{300}+3^{300}+4^{300}>729.24^{100}\)