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Ta có
\(6^{20}=\left(2.3\right)^{20}=2^{20}.3^{20}\)
Mặt khác
\(3^{40}=2^{20}.2^{20}\)
Mà \(2^{20}.2^{30}>2^{20}.2^{20}\)
\(\Rightarrow6^{20}>3^{40}\)
\(6^{20}\) và \(3^{40}\)
\(3^{40}=\left(3^2\right)^{20}=9^{20}\)
vì \(6^{20}< 9^{20}\Rightarrow6^{20}< 3^{40}\)
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
.
Bài 2:
a: \(9^{20}=81^{10}\)
mà 81<9999
nên \(9^{20}< 9999^{10}\)
b: \(9^{20}=3^{40}\)
\(27^{13}=3^{39}\)
mà 40>39
nên \(9^{20}>27^{13}\)
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
VẬY B<A
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
a ) \(\left(-\frac{40}{52}.0,32.\frac{17}{20}\right):\frac{64}{75}\)
= \(\left(-\frac{16}{65}.\frac{17}{20}\right):\frac{64}{75}\)
= \(\left(-\frac{68}{325}\right):\frac{64}{75}\)
= \(\frac{-51}{208}\)
b ) \(-\frac{10}{11}.\frac{8}{9}+\frac{7}{18}.\frac{10}{11}\)
= \(\frac{10}{11}.\left(-\frac{8}{9}+\frac{7}{18}\right)\)
= \(\frac{10}{11}.\left(-\frac{1}{2}\right)\)
= \(\frac{-5}{11}\)
c ) \(\frac{45^{10}.5^{20}}{75^{15}}\)
= \(\frac{5^{10}.3^{20}.5^{20}}{5^{30}.3^{15}}\)
= \(\frac{5^{30}.3^{20}}{5^{30}.3^{15}}\)
= 3 5
= 243
d ) ( - 0,125 ) 3 . 80 4
= -80000
Ta có: 1020 =(102)10 ; 2010
hay 10010 ; 2010
Vì 100 > 20 nên 10010 > 2010
=> 1020 > 2010
mk cũng ko chắc lắm nha bạn
\(10^{20}\) và \(20^{10}\)
\(10^{20}=\left(10^2\right)^{10}=100^{10}\)
vì \(100^{10}>20^{10}\Rightarrow10^{20}>20^{10}\)