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\[1\div 5\]
+
−
.
ln
>
<
×
÷
/
log
≥
≤
(
)
log
x
=
%
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1#5
1#5
1\div 5
1
Use the algorithm method.
0
.
2
5
|
1
.
1
.
0
2
Therefore,
1
÷
5
=
0.2
1\div 5=0.2
1
÷
5
=
0
.
2
.
0.2
0.2
0
.
2
Done
0.2
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