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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 \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
1) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
2) \(3^{21}=3^{20}\cdot3=9^{10}\cdot3\)
\(2^{31}=2^{30}\cdot2=8^{10}\cdot2\)
mà \(9^{10}\cdot3>8^{10}\cdot2\)=> tự viết tiếp
3) đợi chút
430 = (43)10 = 6410 > 4810 = ( 2 . 24 )10 = ( 210 ) . ( 2410 ) > 3 . 2410
=> 230 + 330 + 430 > 3 . 2410
.
\(VT=2^{30}+3^{20}+4^{30}\)
\(=\left(2^3\right)^{10}+\left(3^2\right)^{10}+\left(4^3\right)^{10}\)
\(=8^{10}+9^{10}+64^{10}\)
\(VP=3^{20}+6^{20}+8^{20}\)
\(=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}\)
\(=9^{10}+36^{10}+8^{20}\)
\(=9^{10}+36^{10}+\left(8^2\right)^{10}\)
\(=9^{10}+36^{10}+64^{10}\)
\(\left\{{}\begin{matrix}9^{10}=9^{10}\\64^{10}=64^{10}\\36^{10}>9^{10}\end{matrix}\right.\)
\(\Rightarrow VT< VP\)
\(A=1+3+3^2+3^3+...+3^{2016}\)
\(A=1+3\left(1+3^2+...+3^{2015}\right)\)
\(A=1+3\left(A-3^{2016}\right)\)
\(A=1+3A-3^{2017}\)
\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)
\(A< B\)
a) \(2^{100}=\left(2^2\right)^{50}\)
\(2^2=4< 5\)
\(2^{100}< 5^{50}\)
b) \(4^{30}=\left(4^3\right)^{10}\)
\(4^3=8^2\)
\(4^{30}=8^{20}\)
\(8^{20}=\left(8^2\right)^{10}\)
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 ^-^