a) Ta có:
\(0,3\left(2\right)=0,3222...=0,32+0,00222...\)
Mà: \(0,32+0,00222...>0,32\)
\(\Rightarrow0,3\left(2\right)>0,32\)
b) Ta có:
\(\dfrac{5}{6}=0,8\left(3\right)=0,8333...=0,8+0,0333...\)
\(0,834=0,8+0,034\)
Mà: \(0,0333...< 0,34\)
Nên: \(\dfrac{5}{6}< 0,834\)
a: 0,3(2)=0,3222...>0,32
b: 5/6=0,8(3)=0,83333...<0,834
a) \(0,3\left(2\right)=0,3+\dfrac{2}{90}=\dfrac{29}{90}=\dfrac{145}{450}\)
\(0,32=\dfrac{8}{25}=\dfrac{144}{450}\)
Do \(145>144\Rightarrow\dfrac{145}{450}>\dfrac{144}{450}\)
Vậy \(0,3\left(2\right)>0,32\)
b) \(0,834=\dfrac{417}{500}=\dfrac{1251}{1500}\)
\(\dfrac{5}{6}=\dfrac{1250}{1500}\)
Do \(1251>1250\Rightarrow\dfrac{1251}{1500}>\dfrac{1250}{1500}\)
\(\Rightarrow0,834>\dfrac{5}{6}\)