Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
.
Ta có:
a) 334 = (38)4 = 65614
520 = ( 54)4= 31254
Vì 65614 > 31254 ⇒334 >520
b) 1720 = (174)5 = 835215
Vì 835215 > 715 ⇒1720 > 715.
a) Ta có: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)
b) Ta có : \(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
\(\Rightarrow8192^7>3125^7\)\(\Rightarrow2^{91}>3^{35}\)
c) Ta có: \(99^{20}=99^{2.10}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)
a) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b) Ta có: \(2^{31}=\left(2\frac{31}{21}\right)^{21}=2,7822^{21}< 3^{21}\Rightarrow2^{31}< 3^{21}\)
c) Ta có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(4^{30}=\left(4^3\right)^{10}=64^{10}\)
Lại có: \(3.24^{10}=2.24^{10}+24^{10}\Rightarrow24^{10}< 27^{10}\left(1\right)\)
\(2.24^{10}< 48^{10}< 64^{10}\left(2\right)\)
Từ 1,2 => \(24^{10}+2.24^{10}< 27^{10}+64^{10}\Rightarrow3.24^{10}< 8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3.24^{10}< 3^{30}+2^{30}+4^{30}\)
a, 2225 = 215.15= ( 215)15 = 3276815
3150 = 310.15 = ( 310)15 = 5904915
Dễ thấy 32768 < 59049 nên 2225 < 3150
b)Ta có:
\(17^{20}=17^{4.5}=\left(17^4\right)^5=83521^5>71^5\)
c)Ta có:
\(0,3^{20}=\left(0,3^2\right)^{10}=0,09^{10}< 0,1^{10}\)
d)Ta có:
\(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2}\right)^{40}\)
\(\left(\frac{1}{8}\right)^{13}=\left(\frac{1}{2}\right)^{39}\)
Vì \(\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{2}\right)^{39}\)
nên \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{8}\right)^{13}\)
e)Ta có:
\(3^{21}=3^{20}.3=9^{10}.3\)
\(2^{31}=2^{30}.2=8^{10}.2\)
Vì \(9^{10}.3>8^{10}.2\)
\(\Rightarrow3^{21}>2^{31}\)
a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)
\(7^2=49=7+42\)
mà \(15+2\sqrt{105}< 42\)
nên \(\sqrt{7}+\sqrt{15}< 7\)
b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)
\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)
mà \(2\sqrt{22}< 15+10\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)
9920=(992)10=980110
vì 9801<9999
nên 980110<999910 hay
9920<999910
a) Ta có 9920=(992)10=980110
Vì 980110<999910 nên =>9920<999910
Ta có 231= 2.230=2.(23)10=2.810
321= 3.320=3.(32)10=3.910
Vì 2<3 ;810< 910 nên =>2.810<3.910
hay 231<321
d) ta có 17^20=(17^4)5=83521^5
vì 71^5<83521^5 nên => 71^5<17^20