Giải:
a) Có: \(0,\left(37\right)=0,373737373737...\)
\(0,\left(62\right)=0,626262626262...\)
\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999999...\)
Mà \(0,9999999999999...\simeq1\)
Hay \(0,\left(9\right)=1\)
Vậy \(0,\left(37\right)+0,\left(62\right)=1\).
b) \(0,\left(33\right).3=0,99999...=0,\left(9\right)=1\)
Vậy \(0,\left(33\right).3=1\).
Chúc bạn học tốt!!!
\(a)0,\left(37\right)=0,37373737....\)
\(0,\left(62\right)=0,62626262....\)\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999....\)
Mà \(0,99999999....\simeq1\)
hoặc \(0,\left(9\right)\simeq1\)
\(\Rightarrow0,\left(37\right)+\left(0,62\right)=1\)
\(b)0,\left(33\right).3=1\)
\(\Leftrightarrow0,99999999....=0,\left(9\right)\simeq1\)
\(\Rightarrow0,\left(33\right).3=1\)
Chúc bạn học tốt!
==" ko hỉu 2 người kia giải sao nữa ???
Ta có:
\(0,\left(37\right)=\dfrac{37}{99}\)(cái này vì 0,0(1) = 1/99 nhé, mk khỏi cm)
\(0,\left(62\right)=\dfrac{62}{99}\)
\(\Rightarrow0,\left(37\right)+0,\left(62\right)=\dfrac{37}{99}+\dfrac{62}{99}=\dfrac{99}{99}=1\left(đpcm\right)\)
b, Tương tự:
\(0,\left(33\right)=\dfrac{33}{99}\Rightarrow0,\left(33\right).3=\dfrac{33}{99}.3=1\left(đpcm\right)\)
