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Xét \(A=2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(2^2\right)^{30}=8^{10}+27^{10}+2^{60}\)
\(B=3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}=9^{10}+36^{10}+2^{60}\)
Vì \(8^{10}< 9^{10},27^{10}< 36^{10}\)nên A<B
230 = 23.10= 810
330=33.10=2710
430=43.10=6410
Vế trái = 810 + 2710 + 6410
320=32.10=910
620=62.10=3610
820=82.10=6410
vế phải = 910 + 3610 + 6410
Vì 6410=6410 ; 3610 > 2710 ; 910 > 810
=> vế phải > vế trái
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 \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
a) Ta có : (1/16)10 = [(1/2)4]10 = (1/2)40
Vì (1/2)40 < (1/2)50 nên (1/16)10 < (1/2)50
b) Ta có : 430 = ( 2 . 2)30 = 230 . 230 = (22)15 . (23)10 > 315 . 810 > 3 . 310 .810 = 3 . (3 . 8)10 = 3 .2410
Vậy nên 230 + 330 + 430 > 2410 . 3
Mình chỉ giải thế thôi, còn đâu bn tự làm tiếp
ta có \(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(4^{30}=\left(4^3\right)^{10}=64^{10}\)
ta có \(3^{20}=\left(3^2\right)^{10}=9^{10}\)
\(6^{20}=\left(6^2\right)^{10}=36^{10}\)
\(8^{20}=\left(8^2\right)^{10}=64^{10}\)
\(\Rightarrow2^{30}+3^{30}+4^{30}=8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3^{20}+6^{20}+8^{20}=9^{10}+36^{10}+64^{10}\)
Xét \(8^{10}
\(2^0+2^1+2^2+2^3+...+2^{50}=1+2+2.2+2^2.2+...+2^{49}.2\)
\(=1+2\left(1+2+2^2+2^3+...+2^{49}\right)\)
\(=1+2\left(2^{50}-1\right)\)
\(=1+2^{51}-2\)
\(=2^{51}-1< 2^{51}\)
Vậy \(2^0+2^1+2^2+2^3+...+2^{50}< 2^{51}\)
Ý trc mình ko biết sorry bạn nhiều
T i c k cho mình nha mình mới có 4 điểm, thanks
Ta có:\(2^{30}=\left(2^3\right)^{10}=8^{10}< 9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(3^{30}=3^{20}.3^{10}< 3^{20}.4^{10}=3^{20}.\left(2^2\right)^{10}=3^{20}.2^{20}=\left(3.2\right)^{20}=6^{20}\)
\(4^{30}=4^{20}.4^{10}=4^{20}.\left(2^2\right)^{10}=4^{20}.2^{20}=\left(4.2\right)^{20}=8^{20}\)
nên \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
Ta có: \(2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(4^3\right)^{10}=8^{10}+27^{10}+64^{10}\)
\(3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(8^2\right)^{10}=9^{10}+36^{10}+64^{10}\)
Vì \(8< 9\)\(\Rightarrow8^{10}< 9^{10}\)
mà \(27< 36\)\(\Rightarrow27^{10}< 36^{10}\)
\(\Rightarrow8^{10}+27^{10}< 9^{10}+36^{10}\)
\(\Rightarrow8^{10}+27^{10}+64^{10}< 9^{10}+36^{10}+64^{10}\)
hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
so sánh: 2^30 + 3^30 + 4^30 và 3^20 + 6^20 + 8^20
2^30 = ( 2^3)^10 = 8^ 10
3^30 = (3^3)^10 = 27^10
4^30 = (4^3)^10 = 64^10
3^20 = (3^2)^10 = 9^10
6^20 = (6^2) = 36^10
8^20 = (8^2)^10 = 84^10
vì 9^10 > 8^10
36^10 > 27^10
84^10 > 64^10
=> 2^30 + 3^30 + 4^30 < 3^20 + 6^20 + 8^20
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 ^-^