Question

[Nguyễn Thị Kim Nguyên]

Ssánh

a)\(2^{30}+3^{30}+4^{30}\)và \(3^{20}+6^{20}+8^{20}\)

b) \(1+2^1+2^2+2^3+...+2^{99}\)và \(2^{100}\)

Bạn nào nhanh mjk tjck cho

Hên thì 2

Answer 1

[Hoàng Thủy Tiên]

a, \(2^{30}+3^{30}+4^{30}\) và \(3^{20}+6^{20}+8^{20}\)

Ta có

+) \(2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(4^3\right)^{10}=8^{10}+27^{10}+64^{10}\)

+) \(3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(8^2\right)^{10}=9^{10}+36^{10}+64^{10}\)

Do \(8^{10}+27^{10}+64^{10}< 9^{10}+36^{10}+64^{10}\)

\(\Rightarrow2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)

vậy........

b, \(A=1+2^1+2^2+2^3+......+2^{99}\)

\(2A=2^1+2^2+2^3+2^4+.....+2^{100}\)

\(2A-A=2^{100}-1\)

\(A=2^{100}-1\)

do \(2^{100}-1< 2^{100}\)

\(\Rightarrow1+2^1+2^2+2^3+...+2^{99}< 2^{100}\)

Vậy.......

chúc pn hk tốt ^-^