反殷芳, with real name Billy Chow, is an editor of Wikipedia, who is fond of maths and science.
My calculation for trignometry [ edit ]
sin
31
∘
=
[
2
5
+
5
−
2
(
5
−
1
)
]
{
6
(
5
+
5
)
+
5
−
1
+
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
i
3
+
6
(
5
+
5
)
+
5
−
1
−
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
i
3
}
−
[
2
5
+
5
−
2
(
5
−
1
)
]
{
6
(
5
+
5
)
+
5
−
1
+
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
i
3
−
6
(
5
+
5
)
+
5
−
1
−
[
2
(
5
+
5
)
−
3
(
5
−
1
)
]
i
3
}
i
32
{\displaystyle \sin 31^{\circ }={\tfrac {\left\lbrack 2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\left({\sqrt {5}}-1\right)\right\rbrack \left\lbrace {\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1+\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}+{\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1-\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}\right\rbrace -\left\lbrack 2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\left({\sqrt {5}}-1\right)\right\rbrack \left\lbrace {\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1+\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}-{\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1-\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}\right\rbrace \mathrm {i} }{32}}\,}
Chemical equations included in Chinese junior secondary school chemistry textbooks [ edit ] Note: the chart is in Chinese, and because of the formulae displayed in Wikipedia is not compatible with Chinese characters, the conditions needed in each reaction are NOT WRITTEN IN THE EQUATIONS
初中化学课本涉及的化学方程式列表
化学方程式
反应条件
所在页码
备注
2
N
a
C
l
+
2
H
2
O
=
=
=
=
2
N
a
O
H
+
H
2
↑
+
C
l
2
↑
{\displaystyle {\rm {2NaCl+2H_{2}O{=\!=\!=\!=\!}2NaOH+H_{2}\uparrow +Cl_{2}\uparrow }}}
通电
9A p2
仅仅提及，不要求掌握
C
u
2
(
O
H
)
2
C
O
3
=
=
=
=
2
C
u
O
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {Cu_{2}(OH)_{2}CO_{3}{=\!=\!=\!=\!}2CuO+H_{2}O+CO_{2}\uparrow }}}
2
C
u
O
+
C
=
=
=
=
2
C
u
+
C
O
2
↑
{\displaystyle {\rm {2CuO+C{=\!=\!=\!=\!}2Cu+CO_{2}\uparrow }}}
Δ
高温
9A p2
9A p2, p111
C
u
S
O
4
+
2
N
a
O
H
=
=
=
=
N
a
2
S
O
4
+
C
u
(
O
H
)
2
↓
{\displaystyle {\rm {CuSO_{4}+2NaOH{=\!=\!=\!=\!}Na_{2}SO_{4}+Cu(OH)_{2}\downarrow }}}
9A p7, p21, p100. 9B p75
C
a
C
O
3
+
2
H
C
l
=
=
=
=
C
a
C
l
2
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {CaCO_{3}+2HCl{=\!=\!=\!=\!}CaCl_{2}+H_{2}O+CO_{2}\uparrow }}}
C
a
(
O
H
)
2
+
C
O
2
=
=
=
=
C
a
C
O
3
↓
+
H
2
O
{\displaystyle {\rm {Ca(OH)_{2}+CO_{2}{=\!=\!=\!=\!}CaCO_{3}\downarrow +H_{2}O}}}
9A p7, p113. 9B p73
9A p7, p14, p115, p118. 9B p55 p68
蜡烛 + 氧气
⟶
{\displaystyle \longrightarrow }
点燃
9A p12
初中阶段唯一一个文字表达式
N
a
2
C
O
3
+
2
H
C
l
=
=
=
=
2
N
a
C
l
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {Na_{2}CO_{3}+2HCl{=\!=\!=\!=\!}2NaCl+H_{2}O+CO_{2}\uparrow }}}
9A p20, p94, p130. 9B p74
C
u
(
O
H
)
2
=
=
=
=
C
u
O
+
H
2
O
{\displaystyle {\rm {Cu(OH)_{2}{=\!=\!=\!=\!}CuO+H_{2}O}}}
Δ
9A p21
2
H
g
+
O
2
=
=
=
=
2
H
g
O
{\displaystyle {\rm {2Hg+O_{2}{=\!=\!=\!=\!}2HgO}}}
2
H
g
O
=
=
=
=
2
H
g
+
O
2
↑
{\displaystyle {\rm {2HgO{=\!=\!=\!=\!}2Hg+O_{2}\uparrow }}}
Δ
Δ
9A p26
9A p26, p50
无
初中阶段课本提及的唯一一个产生金属单质的分解反应
4
P
+
5
O
2
=
=
=
=
2
P
2
O
5
{\displaystyle {\rm {4P+5O_{2}{=\!=\!=\!=\!}2P_{2}O_{5}}}}
点燃
9A p27, p92, p100, p129
S
+
O
2
=
=
=
=
S
O
2
{\displaystyle {\rm {S+O_{2}{=\!=\!=\!=\!}SO_{2}}}}
C
+
O
2
=
=
=
=
C
O
2
{\displaystyle {\rm {C+O_{2}{=\!=\!=\!=\!}CO_{2}}}}
3
F
e
+
2
O
2
=
=
=
=
F
e
3
O
4
{\displaystyle {\rm {3Fe+2O_{2}{=\!=\!=\!=\!}Fe_{3}O_{4}}}}
点燃
点燃（\ 高温）
点燃
9A p33, p59, p97, p101
9A p34, p96, p99, p110. 9B p16
9A p34. 9B p22
2
K
M
n
O
4
=
=
=
=
K
2
M
n
O
4
+
M
n
O
2
+
O
2
↑
{\displaystyle {\rm {2KMnO_{4}{=\!=\!=\!=\!}K_{2}MnO_{4}+MnO_{2}+O_{2}\uparrow }}}
2
H
2
O
2
=
=
=
=
2
H
2
O
+
O
2
↑
{\displaystyle {\rm {2H_{2}O_{2}{=\!=\!=\!=\!}2H_{2}O+O_{2}\uparrow }}}
2
K
C
l
O
3
=
=
=
=
2
K
C
l
+
3
O
2
↑
{\displaystyle {\rm {2KClO_{3}{=\!=\!=\!=\!}2KCl+3O_{2}\uparrow }}}
Δ
催化剂（二氧化锰 \ 硫酸铜溶液）
催化剂（二氧化锰）、Δ
9A p37, p100, p102
9A p39, p50, p59
9A p39
N
H
3
⋅
H
2
O
=
=
=
=
N
H
3
↑
+
H
2
O
{\displaystyle {\rm {NH_{3}\cdot H_{2}O{=\!=\!=\!=\!}NH_{3}\uparrow +H_{2}O}}}
无 （书本上） \ 加热
9A p49
仅作了解
H
2
+
C
l
2
=
=
=
=
2
H
C
l
{\displaystyle {\rm {H_{2}+Cl_{2}{=\!=\!=\!=\!}2HCl}}}
点燃
9A p50, p51
初中阶段唯一两个不在氧气中燃烧的反应 之一（另一个在下面）
2
N
a
+
C
l
2
=
=
=
=
2
N
a
C
l
{\displaystyle {\rm {2Na+Cl_{2}{=\!=\!=\!=\!}2NaCl}}}
点燃
9A p55
2
H
2
+
O
2
=
=
=
=
2
H
2
O
{\displaystyle {\rm {2H_{2}+O_{2}{=\!=\!=\!=\!}2H_{2}O}}}
点燃
9A p79, p81, p95, p99, p144
2
H
2
O
=
=
=
=
2
H
2
↑
+
O
2
↑
{\displaystyle {\rm {2H_{2}O{=\!=\!=\!=\!}2H_{2}\uparrow +O_{2}\uparrow }}}
通电
9A p80
初中阶段唯一一个要求掌握的电解反应
3
F
e
+
4
H
2
O
(
g
)
=
=
=
=
F
e
3
O
4
+
4
H
2
{\displaystyle {\rm {3Fe+4H_{2}O(g){=\!=\!=\!=\!}Fe_{3}O_{4}+4H_{2}}}}
高温
9A p81
仅作了解
F
e
+
C
u
S
O
4
=
=
=
=
C
u
+
F
e
S
O
4
{\displaystyle {\rm {Fe+CuSO_{4}{=\!=\!=\!=\!}Cu+FeSO_{4}}}}
9A p91, p93, p97, p101
2
M
g
+
O
2
=
=
=
=
2
M
g
O
{\displaystyle {\rm {2Mg+O_{2}{=\!=\!=\!=\!}2MgO}}}
点燃
9A p94. 9B p22
C
u
O
+
H
2
=
=
=
=
C
u
+
H
2
O
{\displaystyle {\rm {CuO+H_{2}{=\!=\!=\!=\!}Cu+H_{2}O}}}
Δ
9A p97
C
2
H
4
+
3
O
2
=
=
=
=
2
C
O
2
+
2
H
2
O
{\displaystyle {\rm {C_{2}H_{4}+3O_{2}{=\!=\!=\!=\!}2CO_{2}+2H_{2}O}}}
点燃
9A p101(练习)
仅作了解
F
e
2
O
3
+
3
H
2
=
=
=
=
2
F
e
+
3
H
2
O
{\displaystyle {\rm {Fe_{2}O_{3}+3H_{2}{=\!=\!=\!=\!}2Fe+3H_{2}O}}}
高温
9A p101(练习)
仅作了解
C
a
C
O
3
=
=
=
=
C
a
O
+
C
O
2
↑
{\displaystyle {\rm {CaCO_{3}{=\!=\!=\!=\!}CaO+CO_{2}\uparrow }}}
高温
9A p103
2
C
+
O
2
=
=
=
=
2
C
O
{\displaystyle {\rm {2C+O_{2}{=\!=\!=\!=\!}2CO}}}
点燃
9A p110, p141
2
F
e
2
O
3
+
3
C
=
=
=
=
4
F
e
+
3
C
O
2
↑
{\displaystyle {\rm {2Fe_{2}O_{3}+3C{=\!=\!=\!=\!}4Fe+3CO_{2}\uparrow }}}
高温
9A p111
C
O
2
+
C
=
=
=
=
2
C
O
{\displaystyle {\rm {CO_{2}+C{=\!=\!=\!=\!}2CO}}}
高温
9A p111, p136. 9B p16
初中阶段唯一一个明确指出的吸热反应
初中阶段唯一一个高温条件下的化合反应
初中阶段唯一一个二氧化碳转变为一氧化碳的反应
C
a
C
O
3
+
2
H
C
l
=
=
=
=
C
a
C
l
2
+
H
2
C
O
3
{\displaystyle {\rm {CaCO_{3}+2HCl{=\!=\!=\!=\!}CaCl_{2}+H_{2}CO_{3}}}}
9A p113. 9B p73
一般情况下尽量不要使用
H
2
C
O
3
=
=
=
=
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {H_{2}CO_{3}{=\!=\!=\!=\!}H_{2}O+CO_{2}\uparrow }}}
9A p113, p118. 9B p73
H
2
O
+
C
O
2
=
=
=
=
H
2
C
O
3
{\displaystyle {\rm {H_{2}O+CO_{2}{=\!=\!=\!=\!}H_{2}CO_{3}}}}
9A p118
2
C
O
+
O
2
=
=
=
=
2
C
O
2
{\displaystyle {\rm {2CO+O_{2}{=\!=\!=\!=\!}2CO_{2}}}}
点燃
9A p121
C
u
O
+
C
O
=
=
=
=
C
u
+
C
O
2
{\displaystyle {\rm {CuO+CO{=\!=\!=\!=\!}Cu+CO_{2}}}}
Δ
9A p122
C
a
C
O
3
+
2
H
A
c
⟶
C
a
(
A
c
)
2
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {CaCO_{3}+2HAc\longrightarrow Ca(Ac)_{2}+H_{2}O+CO_{2}\uparrow }}}
9A p122
仅作了解（有机反应中箭头代表“生成”）
C
a
O
+
H
2
O
=
=
=
=
C
a
(
O
H
)
2
{\displaystyle {\rm {CaO+H_{2}O{=\!=\!=\!=\!}Ca(OH)_{2}}}}
9A p136. 9B p56
M
g
+
2
H
C
l
=
=
=
=
M
g
C
l
2
+
H
2
↑
{\displaystyle {\rm {Mg+2HCl{=\!=\!=\!=\!}MgCl_{2}+H_{2}\uparrow }}}
9A p136. 9B p10~11
C
H
4
+
2
O
2
=
=
=
=
C
O
2
+
2
H
2
O
{\displaystyle {\rm {CH_{4}+2O_{2}{=\!=\!=\!=\!}CO_{2}+2H_{2}O}}}
点燃
9A p139
C
H
4
⋅
x
H
2
O
+
H
2
O
=
=
=
=
C
O
2
+
(
2
+
x
)
H
2
O
{\displaystyle \mathrm {CH_{4}} \cdot x\mathrm {H_{2}O+H_{2}O{=\!=\!=\!=\!}CO_{2}+} (2+x)\mathrm {H_{2}O} }
点燃
9A p142
仅作了解
C
2
H
5
O
H
+
3
O
2
=
=
=
=
2
C
O
2
+
3
H
2
O
{\displaystyle {\rm {C_{2}H_{5}OH+3O_{2}{=\!=\!=\!=\!}2CO_{2}+3H_{2}O}}}
点燃
9A p144
Z
n
+
H
2
S
O
4
=
=
=
=
Z
n
S
O
4
+
H
2
↑
{\displaystyle {\rm {Zn+H_{2}SO_{4}{=\!=\!=\!=\!}ZnSO_{4}+H_{2}\uparrow }}}
9A p145
4
A
l
+
3
O
2
=
=
=
=
2
A
l
2
O
3
{\displaystyle {\rm {4Al+3O_{2}{=\!=\!=\!=\!}2Al_{2}O_{3}}}}
9B p9, p22
Z
n
+
2
H
C
l
=
=
=
=
Z
n
C
l
2
+
H
2
↑
{\displaystyle {\rm {Zn+2HCl{=\!=\!=\!=\!}ZnCl_{2}+H_{2}\uparrow }}}
F
e
+
2
H
C
l
=
=
=
=
F
e
C
l
2
+
H
2
↑
{\displaystyle {\rm {Fe+2HCl{=\!=\!=\!=\!}FeCl_{2}+H_{2}\uparrow }}}
9B p10~11
9B p10~11
9B p10~11
2
A
l
+
3
C
u
S
O
4
=
=
=
=
A
l
2
(
S
O
4
)
3
+
3
C
u
{\displaystyle {\rm {2Al+3CuSO_{4}{=\!=\!=\!=\!}Al_{2}(SO_{4})_{3}+3Cu}}}
C
u
+
2
A
g
N
O
3
=
=
=
=
C
u
(
N
O
3
)
2
+
2
A
g
{\displaystyle {\rm {Cu+2AgNO_{3}{=\!=\!=\!=\!}Cu(NO_{3})_{2}+2Ag}}}
9B p12
9B p12
F
e
3
O
4
+
4
C
O
=
=
=
=
3
F
e
+
4
C
O
2
{\displaystyle {\rm {Fe_{3}O_{4}+4CO{=\!=\!=\!=\!}3Fe+4CO_{2}}}}
9B p13(练习)
F
e
2
O
3
+
3
C
O
=
=
=
=
2
F
e
+
3
C
O
2
{\displaystyle {\rm {Fe_{2}O_{3}+3CO{=\!=\!=\!=\!}2Fe+3CO_{2}}}}
9B p16, p17, p20
Z
n
C
O
3
=
=
=
=
Z
n
O
+
C
O
2
↑
{\displaystyle {\rm {ZnCO_{3}{=\!=\!=\!=\!}ZnO+CO_{2}\uparrow }}}
2
C
u
2
O
+
C
=
=
=
=
4
C
u
+
C
O
2
↑
{\displaystyle {\rm {2Cu_{2}O+C{=\!=\!=\!=\!}4Cu+CO_{2}\uparrow }}}
800℃
800℃
9B p21(练习)
9B p21(练习)
F
e
+
H
2
S
O
4
=
=
=
=
F
e
S
O
4
+
H
2
↑
{\displaystyle {\rm {Fe+H_{2}SO_{4}{=\!=\!=\!=\!}FeSO_{4}+H_{2}\uparrow }}}
2
A
l
+
3
H
2
S
O
4
=
=
=
=
A
l
2
(
S
O
4
)
3
+
3
H
2
↑
{\displaystyle {\rm {2Al+3H_{2}SO_{4}{=\!=\!=\!=\!}Al_{2}(SO_{4})_{3}+3H_{2}\uparrow }}}
M
g
+
H
2
S
O
4
=
=
=
=
M
g
S
O
4
+
H
2
↑
{\displaystyle {\rm {Mg+H_{2}SO_{4}{=\!=\!=\!=\!}MgSO_{4}+H_{2}\uparrow }}}
9B p22
9B p22
9B p22
F
e
2
O
3
+
6
H
C
l
=
=
=
=
2
F
e
C
l
3
+
3
H
2
O
{\displaystyle {\rm {Fe_{2}O_{3}+6HCl{=\!=\!=\!=\!}2FeCl_{3}+3H_{2}O}}}
F
e
2
O
3
+
3
H
2
S
O
4
=
=
=
=
F
e
2
(
S
O
4
)
3
+
3
H
2
O
{\displaystyle {\rm {Fe_{2}O_{3}+3H_{2}SO_{4}{=\!=\!=\!=\!}Fe_{2}(SO_{4})_{3}+3H_{2}O}}}
9B p54, p68
9B p54, p68
2
N
a
O
H
+
C
O
2
=
=
=
=
N
a
2
C
O
3
+
H
2
O
{\displaystyle {\rm {2NaOH+CO_{2}{=\!=\!=\!=\!}Na_{2}CO_{3}+H_{2}O}}}
9B p56
N
a
O
H
+
H
C
l
=
=
=
=
N
a
C
l
+
H
2
O
{\displaystyle {\rm {NaOH+HCl{=\!=\!=\!=\!}NaCl+H_{2}O}}}
C
a
(
O
H
)
2
+
2
H
C
l
=
=
=
=
C
a
C
l
2
+
2
H
2
O
{\displaystyle {\rm {Ca(OH)_{2}+2HCl{=\!=\!=\!=\!}CaCl_{2}+2H_{2}O}}}
2
N
a
O
H
+
H
2
S
O
4
=
=
=
=
N
a
2
S
O
4
+
2
H
2
O
{\displaystyle {\rm {2NaOH+H_{2}SO_{4}{=\!=\!=\!=\!}Na_{2}SO_{4}+2H_{2}O}}}
9B p60, p68
9B p60
9B p60, p75
A
l
(
O
H
3
)
+
3
H
C
l
=
=
=
=
A
l
C
l
3
+
3
H
2
O
{\displaystyle {\rm {Al(OH_{3})+3HCl{=\!=\!=\!=\!}AlCl_{3}+3H_{2}O}}}
M
g
(
O
H
)
2
+
2
H
A
c
=
=
=
=
M
g
(
A
c
)
2
+
2
H
2
O
{\displaystyle {\rm {Mg(OH)_{2}+2HAc{=\!=\!=\!=\!}Mg(Ac)_{2}+2H_{2}O}}}
9B p65(小题)
9B p65(小题)
2
N
a
O
H
+
S
O
3
=
=
=
=
N
a
2
S
O
4
+
H
2
O
{\displaystyle {\rm {2NaOH+SO_{3}{=\!=\!=\!=\!}Na_{2}SO_{4}+H_{2}O}}}
9B p68
C
a
(
O
H
)
2
+
H
2
S
O
4
=
=
=
=
C
a
S
O
4
+
2
H
2
O
{\displaystyle {\rm {Ca(OH)_{2}+H_{2}SO_{4}{=\!=\!=\!=\!}CaSO_{4}+2H_{2}O}}}
9B p68
N
a
H
C
O
3
+
H
C
l
=
=
=
=
N
a
C
l
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {NaHCO_{3}+HCl{=\!=\!=\!=\!}NaCl+H_{2}O+CO_{2}\uparrow }}}
9B p74
N
a
2
C
O
3
+
C
a
(
O
H
)
2
=
=
=
=
C
a
C
O
3
↓
+
2
N
a
O
H
{\displaystyle {\rm {Na_{2}CO_{3}+Ca(OH)_{2}{=\!=\!=\!=\!}CaCO_{3}\downarrow +2NaOH}}}
9B p74
C
u
S
O
4
+
B
a
C
l
2
=
=
=
=
B
a
S
O
4
↓
+
C
u
C
l
2
{\displaystyle {\rm {CuSO_{4}+BaCl_{2}{=\!=\!=\!=\!}BaSO_{4}\downarrow +CuCl_{2}}}}
9B p74
K
2
C
O
3
+
H
2
S
O
4
=
=
=
=
K
2
S
O
4
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {K_{2}CO_{3}+H_{2}SO_{4}{=\!=\!=\!=\!}K_{2}SO_{4}+H_{2}O+CO_{2}\uparrow }}}
B
a
(
N
O
3
)
2
+
H
2
S
O
4
=
=
=
=
B
a
S
O
4
↓
+
2
H
N
O
3
{\displaystyle {\rm {Ba(NO_{3})_{2}+H_{2}SO_{4}{=\!=\!=\!=\!}BaSO_{4}\downarrow +2HNO_{3}}}}
9B p75
9B p75
C
a
C
O
3
+
H
2
O
+
C
O
2
=
=
=
=
C
a
(
H
C
O
3
)
2
{\displaystyle {\rm {CaCO_{3}+H_{2}O+CO_{2}{=\!=\!=\!=\!}Ca(HCO_{3})_{2}}}}
C
a
(
H
C
O
3
)
2
=
=
=
=
C
a
C
O
3
↓
+
H
2
O
+
C
O
2
↑
{\displaystyle {\rm {Ca(HCO_{3})_{2}{=\!=\!=\!=\!}CaCO_{3}\downarrow +H_{2}O+CO_{2}\uparrow }}}
9B p77
9B p77
(
N
H
4
)
2
S
O
4
+
C
a
(
O
H
)
2
=
=
=
=
C
a
S
O
4
+
2
H
2
O
+
2
N
H
3
↑
{\displaystyle {\rm {(NH_{4})_{2}SO_{4}+Ca(OH)_{2}{=\!=\!=\!=\!}CaSO_{4}+2H_{2}O+2NH_{3}\uparrow }}}
2
N
H
4
C
l
+
C
a
(
O
H
)
2
=
=
=
=
C
a
C
l
2
+
2
H
2
O
+
2
N
H
3
↑
{\displaystyle {\rm {2NH_{4}Cl+Ca(OH)_{2}{=\!=\!=\!=\!}CaCl_{2}+2H_{2}O+2NH_{3}\uparrow }}}
N
H
4
H
C
O
3
=
=
=
=
H
2
O
+
N
H
3
↑
+
C
O
2
↑
{\displaystyle {\rm {NH_{4}HCO_{3}{=\!=\!=\!=\!}H_{2}O+NH_{3}\uparrow +CO_{2}\uparrow }}}
9B p83
9B p83
9B p83
C
6
H
12
O
6
+
6
O
2
=
=
=
=
6
C
O
2
+
6
H
2
O
{\displaystyle {\rm {C_{6}H_{12}O_{6}+6O_{2}{=\!=\!=\!=\!}6CO_{2}+6H_{2}O}}}
酶
9B p93
在平面直角坐标系中：
斜截式：若一次函数斜率为k，y轴上截距为b，则：
y
=
k
x
+
b
{\displaystyle y=kx+b}
点斜式：若一次函数斜率为k，图像经过点
(
x
0
,
y
0
)
{\displaystyle (x_{0},y_{0})}
y
−
y
0
=
k
(
x
−
x
0
)
{\displaystyle y-y_{0}=k(x-x_{0})}
两点式：若一次函数图像经过点
(
x
1
,
y
1
)
{\displaystyle (x_{1},y_{1})}
(
x
2
,
y
2
)
{\displaystyle (x_{2},y_{2})}
y
−
y
1
x
−
x
1
=
y
2
−
y
1
x
2
−
x
1
{\displaystyle {\frac {y-y_{1}}{x-x_{1}}}={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}
截距式：若一次函数x轴上截距为a，y轴上截距为b，则：
x
a
+
y
b
=
1
{\displaystyle {\frac {x}{a}}+{\frac {y}{b}}=1}
一般式：
A
x
+
B
y
+
C
=
0
{\displaystyle Ax+By+C=0}
一般式：
y
=
a
x
2
+
b
x
+
c
{\displaystyle y=ax^{2}+bx+c}
顶点式：若二次函数顶点为
(
h
,
k
)
{\displaystyle (h,k)}
y
=
a
(
x
−
h
)
2
+
k
{\displaystyle y=a(x-h)^{2}+k}
两点式：若二次函数图像交x轴于
(
x
1
,
0
)
{\displaystyle (x_{1},0)}
(
x
2
,
0
)
{\displaystyle (x_{2},0)}
y
=
a
(
x
−
x
1
)
(
x
−
x
2
)
{\displaystyle y=a(x-x_{1})(x-x_{2})}
若有点
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
B
(
x
2
,
y
2
)
{\displaystyle B(x_{2},y_{2})}
A
B
=
(
x
1
−
x
2
)
2
+
(
y
1
−
y
2
)
2
{\displaystyle AB={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}}
若有点
P
(
x
0
,
y
0
)
{\displaystyle P(x_{0},y_{0})}
l
:
A
x
+
B
y
+
C
=
0
{\displaystyle l:Ax+By+C=0}
d
=
|
A
x
0
+
B
y
0
+
C
|
A
2
+
B
2
{\displaystyle d={\frac {\left\vert Ax_{0}+By_{0}+C\right\vert }{\sqrt {A^{2}+B^{2}}}}}
若有平行线
l
1
:
y
=
k
x
+
b
1
{\displaystyle l_{1}:y=kx+b_{1}}
l
2
:
y
=
k
x
+
b
2
{\displaystyle l_{2}:y=kx+b_{2}}
d
=
|
b
1
−
b
2
|
k
2
+
1
{\displaystyle d={\frac {\left\vert b_{1}-b_{2}\right\vert }{\sqrt {k^{2}+1}}}}
若直线l1 斜率为k1 ，直线l2 斜率为k2 ，且
k
1
k
2
≠
−
1
{\displaystyle k_{1}k_{2}\neq -1}
1 到l2 的角为：
tan
θ
=
k
2
−
k
1
1
+
k
1
k
2
{\displaystyle \tan \theta ={\frac {k_{2}-k_{1}}{1+k_{1}k_{2}}}}
1 与l2 的夹角为：
tan
θ
=
|
k
2
−
k
1
1
+
k
1
k
2
|
{\displaystyle \tan \theta =\left\vert {\frac {k_{2}-k_{1}}{1+k_{1}k_{2}}}\right\vert }
注：
直线l1 到l2 的角 指从直线l1 逆时针 旋转到l2 所需的角度（
0
⩽
θ
⩽
180
∘
{\displaystyle 0\leqslant \theta \leqslant 180^{\circ }}
直线l1 与l2 的夹角 指直线l1 与l2 相交时所成的锐角或直角 （
0
⩽
θ
⩽
90
∘
{\displaystyle 0\leqslant \theta \leqslant 90^{\circ }}
若l1 ⊥l2 ，则：
k
1
k
2
=
−
1
{\displaystyle k_{1}k_{2}=-1}
若有点
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
B
(
x
2
,
y
2
)
{\displaystyle B(x_{2},y_{2})}
M
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle M({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})}
若有三点点
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
B
(
x
2
,
y
2
)
{\displaystyle B(x_{2},y_{2})}
C
(
x
3
,
y
3
)
{\displaystyle C(x_{3},y_{3})}
G
(
x
1
+
x
2
+
x
3
3
,
y
1
+
y
2
+
y
3
3
)
{\displaystyle G({\frac {x_{1}+x_{2}+x_{3}}{3}},{\frac {y_{1}+y_{2}+y_{3}}{3}})}
若
⊙
O
{\displaystyle \odot O}
(
a
,
b
)
{\displaystyle (a,b)}
(
x
−
a
)
2
+
(
y
−
b
)
2
=
r
2
{\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}}
直线
l
:
A
x
+
B
y
+
C
=
0
{\displaystyle l:Ax+By+C=0}
(
x
−
a
)
2
+
(
y
−
b
)
2
=
r
2
{\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}}
Δ
{\displaystyle \Delta }
d
=
|
A
a
+
B
b
+
C
|
A
2
+
B
2
{\displaystyle d={\frac {\left\vert Aa+Bb+C\right\vert }{\sqrt {A^{2}+B^{2}}}}}
d
>
r
⇔
{\displaystyle d>r\Leftrightarrow }
⇔
Δ
<
0
{\displaystyle \Leftrightarrow \Delta <0}
d
=
r
⇔
{\displaystyle d=r\Leftrightarrow }
⇔
Δ
=
0
{\displaystyle \Leftrightarrow \Delta =0}
d
<
r
⇔
{\displaystyle d<r\Leftrightarrow }
⇔
Δ
<
0
{\displaystyle \Leftrightarrow \Delta <0}
f (x )=sin(kx )cos(jx )的周期[ edit ] 1月17日手稿（已与戴） 显然
f
(
x
)
=
sin
k
x
cos
j
x
=
1
2
[
sin
(
k
+
j
)
x
+
sin
(
k
−
j
)
x
]
{\displaystyle f(x)=\sin kx\cos jx={\frac {1}{2}}\left[\sin(k+j)x+\sin(k-j)x\right]}
令
p
=
k
+
j
{\displaystyle p=k+j}
q
=
k
−
j
{\displaystyle q=k-j}
f
(
x
)
=
1
2
(
sin
p
x
+
sin
q
x
)
{\displaystyle f(x)={\frac {1}{2}}\left(\sin px+\sin qx\right)}
⇒
f
(
x
+
2
π
n
)
=
1
2
[
sin
(
p
x
+
2
p
π
n
)
+
sin
(
q
x
+
2
q
π
n
)
]
{\displaystyle \Rightarrow f\left(x+{\frac {2\pi }{n}}\right)={\frac {1}{2}}\left[\sin \left(px+{\frac {2p\pi }{n}}\right)+\sin \left(qx+{\frac {2q\pi }{n}}\right)\right]}
∵sin(x )的周期为
2
π
{\displaystyle 2\pi }
2
p
π
n
{\displaystyle {\frac {2p\pi }{n}}}
2
q
π
n
{\displaystyle {\frac {2q\pi }{n}}}
2
π
{\displaystyle 2\pi }
p
n
,
q
n
∈
Z
{\displaystyle {\frac {p}{n}},{\frac {q}{n}}\in \mathbb {Z} }
情况一：若
p
,
q
∈
Z
{\displaystyle p,q\in \mathbb {Z} }
n
∣
p
,
n
∣
q
{\displaystyle n\mid p,\ n\mid q}
T
{\displaystyle T}
n
=
gcd
(
|
p
|
,
|
q
|
)
{\displaystyle n=\gcd(\left\vert p\right\vert ,\left\vert q\right\vert )}
T
=
2
π
gcd
(
|
k
+
j
|
,
|
k
−
j
|
)
{\displaystyle T={\frac {2\pi }{\gcd(\left\vert k+j\right\vert ,\left\vert k-j\right\vert )}}}
情况二：若
p
∉
Z
∪
q
∉
Z
{\displaystyle p\notin \mathbb {Z} \cup q\notin \mathbb {Z} }
1) 若
(
p
∈
R
∖
Q
)
∪
(
q
∈
R
∖
Q
)
⇒
f
(
x
)
{\displaystyle \left(p\in \mathbb {R} \setminus \mathbb {Q} \right)\cup \left(q\in \mathbb {R} \setminus \mathbb {Q} \right)\Rightarrow f(x)}
2) 若
(
p
∈
Q
)
∩
(
q
∈
Q
)
{\displaystyle \left(p\in \mathbb {Q} \right)\cap \left(q\in \mathbb {Q} \right)}
p
=
a
1
b
1
{\displaystyle p={\frac {a_{1}}{b_{1}}}}
q
=
a
2
b
2
{\displaystyle q={\frac {a_{2}}{b_{2}}}}
a
1
,
a
2
,
b
1
,
b
2
∈
Z
,
gcd
(
a
1
,
b
1
)
=
gcd
(
a
2
,
b
2
)
=
1
{\displaystyle a_{1},a_{2},b_{1},b_{2}\in \mathbb {Z} ,\ \gcd(a_{1},b_{1})=\gcd(a_{2},b_{2})=1}
则
a
1
b
1
n
,
a
2
b
2
n
∈
Z
⇒
1
a
1
n
,
1
b
1
n
∈
Z
{\displaystyle {\frac {a_{1}}{b_{1}n}},\ {\frac {a_{2}}{b_{2}n}}\in \mathbb {Z} \Rightarrow {\frac {1}{a_{1}n}},\ {\frac {1}{b_{1}n}}\in \mathbb {Z} }
易知
n
=
1
lcm
(
|
b
1
|
,
|
b
2
|
)
⇒
T
=
2
lcm
(
|
b
1
|
,
|
b
2
|
)
π
{\displaystyle n={\frac {1}{\operatorname {lcm} (\left\vert b_{1}\right\vert ,\left\vert b_{2}\right\vert )}}\Rightarrow T=2\operatorname {lcm} (\left\vert b_{1}\right\vert ,\left\vert b_{2}\right\vert )\pi }
[ edit ] 首先要明白这样一个出题的逻辑：命题者肯定是先列出一个不那么复杂的数列关系（譬如某多项式倒数呈等差数列），然后将其进行处理，使其变得复杂。因此，我们宜从结果开始，倒推出这一方法适用的题型。
{
1
a
n
−
x
}
{\displaystyle \left\{{\frac {1}{a_{n}-x}}\right\}}
⇐
1
a
n
+
1
−
x
−
1
a
n
−
x
=
d
{\displaystyle \Leftarrow {\frac {1}{a_{n+1}-x}}-{\frac {1}{a_{n}-x}}=d}
⇐
a
n
+
1
=
(
x
d
+
1
)
a
n
−
x
2
d
d
a
n
−
x
d
+
1
{\displaystyle \Leftarrow a_{n+1}={\frac {(xd+1)a_{n}-x^{2}d}{da_{n}-xd+1}}}
令
m
1
=
x
d
+
1
{\displaystyle m_{1}=xd+1}
n
1
=
−
x
2
d
{\displaystyle n_{1}=-x^{2}d}
m
2
=
d
{\displaystyle m_{2}=d}
n
2
=
−
x
d
+
1
{\displaystyle n_{2}=-xd+1}
a
n
+
1
=
m
1
a
n
+
n
1
m
2
a
n
+
n
2
{\displaystyle a_{n+1}={\frac {m_{1}a_{n}+n_{1}}{m_{2}a_{n}+n_{2}}}}
x
{\displaystyle x}
a
n
+
1
,
a
n
{\displaystyle a_{n+1},\ a_{n}}
方程
cos
x
=
x
{\displaystyle \cos x=x}
x
=
π
2
exp
[
1
π
∫
0
1
arctan
(
π
x
+
2
)
ln
(
1
−
x
2
+
1
x
)
x
x
2
ln
2
(
1
−
x
2
+
1
x
)
−
π
x
−
1
x
d
x
]
{\displaystyle x={\frac {\pi }{2}}\exp \left[{\frac {1}{\pi }}\int _{0}^{1}{\frac {\arctan {\tfrac {(\pi x+2)\ln \left({\tfrac {{\sqrt {1-x^{2}}}+1}{x}}\right)x}{x^{2}\ln ^{2}\left({\tfrac {{\sqrt {1-x^{2}}}+1}{x}}\right)-\pi x-1}}}{x}}\,dx\right]}
求解：
f
(
x
)
+
f
(
x
)
f
′
(
x
)
=
x
{\displaystyle f(x)+f(x)f'(x)=x}
计算
(
2
+
i
)
1
+
i
{\displaystyle (2+i)^{1+i}}
电脑上有一种输入法，其中的字符满足如下规则：同一个拼音中，若选择一个字打出，下一次输入同样拼音时，该字将被置顶。现输入一特定拼音，选择打出其中第
n
{\displaystyle n}
![{\displaystyle \sin 31^{\circ }={\tfrac {\left\lbrack 2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\left({\sqrt {5}}-1\right)\right\rbrack \left\lbrace {\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1+\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}+{\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1-\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}\right\rbrace -\left\lbrack 2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\left({\sqrt {5}}-1\right)\right\rbrack \left\lbrace {\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1+\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}-{\sqrt[{3}]{{\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1-\left\lbrack {\sqrt {2\left(5+{\sqrt {5}}\right)}}-{\sqrt {3}}\left({\sqrt {5}}-1\right)\right\rbrack \mathrm {i} }}\right\rbrace \mathrm {i} }{32}}\,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/23f200daf20c11090d3b29560acfee07aab57b72)
![{\displaystyle f(x)=\sin kx\cos jx={\frac {1}{2}}\left[\sin(k+j)x+\sin(k-j)x\right]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/9b18f896ff280abc1c92e145f80296b2f8c9aafd)
![{\displaystyle \Rightarrow f\left(x+{\frac {2\pi }{n}}\right)={\frac {1}{2}}\left[\sin \left(px+{\frac {2p\pi }{n}}\right)+\sin \left(qx+{\frac {2q\pi }{n}}\right)\right]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/db69f92377a8be25c14cd5b7644bcc61d46c57c7)
![{\displaystyle x={\frac {\pi }{2}}\exp \left[{\frac {1}{\pi }}\int _{0}^{1}{\frac {\arctan {\tfrac {(\pi x+2)\ln \left({\tfrac {{\sqrt {1-x^{2}}}+1}{x}}\right)x}{x^{2}\ln ^{2}\left({\tfrac {{\sqrt {1-x^{2}}}+1}{x}}\right)-\pi x-1}}}{x}}\,dx\right]}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/00d7e33d14529266aa269637314ac3907e0a9727)
