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\[1\div 3\]
+
−
.
ln
>
<
×
÷
/
log
≥
≤
(
)
log
x
=
%
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1#3
1#3
1\div 3
1
Use the algorithm method.
0
.
3
3
3
3
3
3
3
|
1
.
9
1
0
9
1
0
9
1
0
9
1
0
9
1
0
9
1
2
Therefore,
1
÷
3
≈
0.333333
1\div 3\approx0.333333
1
÷
3
≈
0
.
3
3
3
3
3
3
.
0.333333
0.333333
0
.
3
3
3
3
3
3
Done
0.33333333333333
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