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User:Luosiji
20
x
2
+
2
y
2
−
5
=
0
{\displaystyle 20x^{2}+2y^{2}-5=0\!}
36
x
2
+
68
x
y
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36
y
2
−
136
x
−
144
y
+
139
=
0
{\displaystyle 36x^{2}+68xy+36y^{2}-136x-144y+139=0\!}
y
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12
=
−
4
5
(
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x
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+
y
12
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9
4
)
2
3
+
2
{\displaystyle y-{\frac {|x|}{12}}=-{\frac {4}{5}}{\sqrt[{3}]{\left(\left|x\right|+{\frac {y}{12}}-{\frac {9}{4}}\right)^{2}}}+2\!}
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x
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6
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2
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6
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9
4
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−
1
)
2
3
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6
5
{\displaystyle y-{\frac {|x|}{6}}={\frac {1}{2}}{\sqrt[{3}]{\left(\left|\left|x\right|+{\frac {y}{6}}-{\frac {9}{4}}\right|-1\right)^{2}}}-{\frac {6}{5}}\!}
x
2
3
±
9
−
x
2
{\displaystyle {\sqrt[{3}]{x^{2}}}\pm {\sqrt {9-x^{2}}}\!}
∑
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0
n
∑
j
=
1
m
(
m
j
)
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m
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j
=
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n
+
1
)
m
{\displaystyle \sum _{i=0}^{n}\sum _{j=1}^{m}{m \choose j}{i}^{m-j}={\left(n+1\right)}^{m}}
∑
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0
n
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i
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1
)
m
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i
m
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n
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1
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m
{\displaystyle \sum _{i=0}^{n}{\left(i+1\right)}^{m}-{i}^{m}={\left(n+1\right)}^{m}}
E
[
T
]
≤
⌊
log
n
m
⌋
+
1
+
m
−
1
m
n
n
−
1
.
{\displaystyle E[T]\leq \lfloor \log _{n}m\rfloor +1+{\frac {m-1}{m}}{\frac {n}{n-1}}.}