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a/ Ta có: 2100 = 22.50 = (22)50 = 450
550 = 550
Vì 4 < 5 nên 450 < 550
Vậy 2100 < 550
b/ Ta có: 430 = (22)30 = 260
820 = (23)20 = 260
Vì 260 = 260
Nên 430 = 820
\(a.\:2^{100}=\left(2^2\right)^{50}=4^{50}< 5^{50}\)
\(b.\:\left\{{}\begin{matrix}4^{30}=64^{10}\\8^{20}=64^{10}\end{matrix}\right.\Rightarrow4^{30}=8^{20}\)
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: \(25^{15}=\left(5^2\right)^{15}=5^{30}\)
\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}\)\(=2^{30}.3^{30}=6^{30}\)
Vì \(5^{30}< 6^{30}\)nên \(25^{15}< 8^{10}.3^{30}\)
b) Ta có: \(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(2^{30}< 3^{30}\)nên \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)hay \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
_Học tốt_
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 ^-^
a)Ta có:\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Vì \(27^{10}>25^{10}\Rightarrow3^{30}>5^{20}\)
Do 27>25 nên \(27^{10}>25^{10}\)\(hay\) \(3^{30}>5^{20}\)
còn câu b thì mk chưa tính ra
a)27^11=(3^3)^11=3^33
81^8=(3^4)8=3^32
vì 3^33>3^32 nên 27^11>81^8
b)ko biết làm chỉ biết 3^150>2^225
c)27^50=27^5x10=(27^5)^10=14348907^10
240^30=240^3x10=(240^3)^10=13824000^10
suy ra 27^50>240^30
a) Ta có: \(27^{11}=\left(3^3\right)^{^{11}}=3^{3.11}=3^{33}\)
\(81^8=\left(3^4\right)^{^8}=3^{4.8}=3^{32}\)
Vì \(3^{33}>3^{32}\)
nên \(27^{11}>81^8\)
b) Ta có: \(3^{150}=3^{2.75}=\left(3^2\right)^{^{75}}=9^{75}\)
\(2^{225}=2^{3.75}=\left(2^3\right)^{^{75}}=8^{75}\)
vì \(9^{75}>8^{75}\)
nên \(3^{150}>2^{225}\)
c) Ta có:
\(\frac{27^{50}}{240^{30}}=\frac{27^{30}.27^{20}}{240^{30}}=\frac{3^{30}.3^{30}.3^{30}.3^{20}.3^{20}.2^{20}}{3^{30}.80^{30}}\)
\(=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{^{30}}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì \(\frac{81^{30}}{80^{30}}>1\)\(\Rightarrow\frac{27^{50}}{240^{30}}>1\)\(\Rightarrow27^{50}>240^{30}\)
a) Ta có :
\(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
Vì 333 > 332
=> 2711 > 818
b) Ta có:
\(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
Vì 875 < 975
=> 2225 < 3150
Thôi còn lại bn tự làm nốt nha . Nhìn mà nản !!
a) \(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
333 > 332 => 2711 > 818
b) \(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
875 < 975 => 2225 < 3150
c) \(\hept{\begin{cases}2^{500}=\left(2^5\right)^{100}=32^{100}\\5^{200}=\left(5^2\right)^{100}=25^{100}\end{cases}}\)
32100 > 25100 => 2500 > 5200
d) \(\hept{\begin{cases}625^5=\left(5^4\right)^5=5^{20}\\125^7=\left(5^3\right)^7=5^{21}\end{cases}}\)
520 < 521 => 6255 < 1257
e) \(\hept{\begin{cases}5^{100}=\left(5^4\right)^{25}=625^{25}\\8^{75}=\left(8^3\right)^{25}=512^{25}\end{cases}}\)
62525 > 51225 => 5100 > 875
f) \(2^{16}=2^3\cdot2^{13}=8\cdot2^{13}\)
7 < 8 => 7.213 < 8.213 => 7.213 < 216
g) Ta có \(\frac{27^{50}}{240^{30}}=\frac{\left(3^3\right)^{50}}{3^{30}\cdot80^{30}}=\frac{3^{150}}{3^{30}\cdot80^{30}}=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{30}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì 8130 > 8030 => 8130/8030 > 1 => 2750/24030 > 1 => 2750 > 24030
h) Ta có \(\hept{\begin{cases}63^9< 64^9=\left(2^6\right)^9=2^{54}\left(1\right)\\16^{14}=\left(2^4\right)^{14}=2^{56}< 17^{14}\left(2\right)\end{cases}}\)
Từ (1) và (2) => 639 < 254 < 256 < 1714
=> 639 < 1714
a) \(2^{100}=\left(2^2\right)^{50}\)
\(2^2=4< 5\)
\(2^{100}< 5^{50}\)
b) \(4^{30}=\left(4^3\right)^{10}\)
\(4^3=8^2\)
\(4^{30}=8^{20}\)
\(8^{20}=\left(8^2\right)^{10}\)
2100 và 550
Ta có :
2100 = (22)50 = 450
Vì 450 < 550 nên 2100 < 550
430 và 820
Ta có :
430 = (43)10 = 6410
820 = (82)10 = 8110
Vì 6410 < 8110 nên 430 < 820