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290=(25)18=3218
536=(52)18=2518
Vì 32>25 nên 3218>2518
=>290>536
b,15=\(\sqrt{225}\) <\(\sqrt{235}\)
=> 15<\(\sqrt{235}\)
c, Ta có: \(\dfrac{1}{3}=\dfrac{13}{39}\)
vì 38<39
nên \(\dfrac{13}{38}>\dfrac{13}{39}\)
a) 290= (210)9mà 210=(25)2
536= (54)9mà 54=(52)2
Do 25>52nên 290>536
a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Do : \(8^{100}< 9^{100}\)
=> \(2^{300}< 3^{200}\)
b) Do \(\dfrac{13}{38}>\dfrac{13}{39}\)
Mà : \(\dfrac{13}{39}=\dfrac{1}{3}\)
=> \(\dfrac{13}{38}>\dfrac{1}{3}\)
c)Do : \(\sqrt{235}>\sqrt{225}\)
Mà : \(\sqrt{225}=15\)
=> \(\sqrt{235}>15\)
a) Ta có:
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Ta thấy 8<9 suy ra \(8^{100}< 9^{100}\)
Vậy \(2^{300}< 3^{200}\)
\(a,\dfrac{1997}{1996}>1>\dfrac{1996}{1997}\\ b,\dfrac{3}{5}< 1< \dfrac{15}{13}\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
a) \(\dfrac{13}{38}\) và \(\dfrac{1}{3}\)
\(\dfrac{1}{3}\) = \(\dfrac{13}{39}\) < \(\dfrac{13}{38}\)
=> \(\dfrac{13}{38}>\dfrac{1}{3}\)
b)\(\sqrt{235}\) và 15
15 = \(\sqrt{225}\) < \(\sqrt{235}\) ( vì 225 < 235)
=> \(\sqrt{235}>15\)
tick mình nha
=>
a, Ta có:
\(\dfrac{13}{38}\)=\(\dfrac{39}{114}\) ; \(\dfrac{1}{3}\)=\(\dfrac{38}{114}\)
Vì 38 < 39 ⇒ \(\dfrac{39}{114}>\dfrac{38}{114}\)
Vay \(\dfrac{13}{38}>\dfrac{1}{3}\)
b, Goi \(\sqrt{235}\)= a ⇒ 235 = \(a^2\)
Ta có : 15^2= 225
Vì 235 > 225 nên a^2 > 15^2
Vay \(\sqrt{235}\)>15