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210 +320+420 =( 22)10 + (3^2)^10 + (4^2)^10 = 4^10 +9^10+16^10
210 +320+420 =( 22)10 + (3^2)^10 + (4^2)^10 = 4^10 +9^10+16^10
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c) Đặt \(A=2^0+2^1+2^2+...+2^{50}\)
\(\Leftrightarrow2A=2^1+2^2+2^3...+2^{51}\)
\(\Leftrightarrow2A-A=2^1+2^2+2^3...+2^{51}\)\(-2^0-2^1-2^2-...-2^{50}\)
\(\Leftrightarrow A=2^{51}-2^0=2^{51}-1< 2^{51}\)
Vậy \(2^0+2^1+2^2+...+2^{50}< 2^{51}\)
a)Ta có: \(\hept{\begin{cases}2^{30}=\left(2^3\right)^{10}=8^{10}\\3^{30}=\left(3^3\right)^{10}=27^{10}\\4^{30}=\left(2^2\right)^{30}=2^{60}\end{cases}}\)và \(\hept{\begin{cases}3^{20}=\left(3^2\right)^{10}=9^{10}\\6^{20}=\left(6^2\right)^{10}=36^{10}\\8^{20}=\left(2^3\right)^{20}=2^{60}\end{cases}}\)
Mà \(8^{10}< 9^{10}\); \(27^{10}< 36^{10}\);\(2^{60}=2^{60}\)nên
\(8^{10}+27^{10}+2^{60}< 9^{10}+36^{10}+2^{60}\)
hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
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Ta có:
3.24^10=3^11.4^15
=> 4^30=4^15.4^15
4^15>3^11 (vì phần nguyên bé và mũ cũng bé nên ta có:4^15>3^11)
=>3.24^10<<4^30<<<2^30+3^20+4^30
\(4^{30}=2^{30}.2^{30}=\left(2^3\right)^{10}.\left(2^2\right)^{15}=8^{10}.4^{15}>8^{10}.3^{15}>8^{10}.3^{11}\)
\(=8^{10}.3^{10}.3=\left(8.3\right)^{10}.3=3.24^{10}\)
=>2^30+... >3.24^10
tick nhé(bn nói rồi mà)