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a: \(0.2=\dfrac{2}{10}\)
10>7
=>\(\dfrac{2}{10}< \dfrac{2}{7}\)
=>\(\dfrac{2}{7}>0.2\)
b: \(-\dfrac{1^5}{6}=\dfrac{-1}{6}=\dfrac{-3}{18}\)
\(\dfrac{8}{-9}=-\dfrac{16}{18}\)
mà -3>-16
nên \(-\dfrac{1^5}{6}>\dfrac{8}{-9}\)
c: \(\dfrac{2017}{2016}>1\)
\(1>\dfrac{2017}{2018}\)
Do đó: \(\dfrac{2017}{2016}>\dfrac{2017}{2018}\)
d: \(-\dfrac{249}{333}=\dfrac{-249:3}{333:3}=\dfrac{-83}{111}\)
e: \(\dfrac{5^1}{3}=\dfrac{5}{3}=\dfrac{15}{9}\)
\(\dfrac{4^8}{9}=\dfrac{65536}{9}\)
mà 15<65536
nên \(\dfrac{5^1}{3}< \dfrac{4^8}{9}\)
f: 13,589<13,612
ta có :
\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)
mà \(5^{2017}>5^{2016}\)
\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)
\(\Rightarrow\)\(5^{2017}>25^{1008}\)
có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)
mà \(=25^{1008}\times5\)> \(25^{1008}\)
nên \(5^{2017}>25^{1008}\)
a: Ta có: \(A=2018^2-2017^2=2018+2017\)
\(B=2017^2-2016^2=2017+2016\)
mà 2018>2016
nên A>B
`a,` Ta có:
`199/198 > 1`
`2017/2018 < 1`
`-> 2017/2018 < 1 < 199/198`
Vậy: `2017/2018 <, 199/198`
`b,` Ta có:
`-12/48 = -1/4`
`25/(-101)> 25/(-100) = -1/4 `
`-> -12/48 < 25/(-101)`
Vậy: `-12/48 < 25/(-101)`
a: \(\dfrac{199}{198}>\dfrac{198}{198}=1;1=\dfrac{2018}{2018}>\dfrac{2017}{2018}\)
Do đó: \(\dfrac{199}{198}>\dfrac{2017}{2018}\)
b: \(\dfrac{12}{48}=\dfrac{1}{4};\dfrac{1}{4}=\dfrac{25}{100}>\dfrac{25}{101}\)
=>\(\dfrac{12}{48}>\dfrac{25}{101}\)
=>\(-\dfrac{12}{48}< -\dfrac{25}{101}\)