Bài 1 : So Sánh
A = 4 * (3 ^ 2 +1) * (3 ^ 4 + 1 ) * (3 ^ 8 + 1) * . .. * (3 ^ 64 + 1 )
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1:
a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)
b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)
nên 2 p/s ko bằng nhau
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^{128}-1\right)< 3^{128}-1=Q\)
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1\\ P=\dfrac{3^{128}-1}{2}< Q=3^{218}-1\)
\(S=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(\left(3^2-1\right)S=4\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(8S=4\cdot\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(2S=\left(3^8-1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
...
\(2S=3^{128}-1\)
Vậy S < 3128 - 1