Knowledge

User:反殷芳

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反殷芳, with real name Billy Chow, is an editor of Knowledge, who is fond of maths and science.

My calculation for trignometry

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{{\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{{\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{{\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{{\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}}\,}

My chemistry

Chemical equations included in Chinese junior secondary school chemistry textbooks

Note: the chart is in Chinese, and because of the formulae displayed in Knowledge is not compatible with Chinese characters, the conditions needed in each reaction are NOT WRITTEN IN THE EQUATIONS and are shown in the column next to it.

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 } 二氧化碳 + 水 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

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}}} 9B p93

解析几何基本公式

在平面直角坐标系中:

  • 一次函数有以下形式:
  1. 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}
  2. 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})} ,则:
    y y 0 = k ( x x 0 ) {\displaystyle y-y_{0}=k(x-x_{0})}
  3. 两点式:若一次函数图像经过点 ( x 1 , y 1 ) {\displaystyle (x_{1},y_{1})} 和点 ( x 2 , y 2 ) {\displaystyle (x_{2},y_{2})} ,则:
    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}}}}
  4. 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}
  5. 一般式: A x + B y + C = 0 {\displaystyle Ax+By+C=0}
  • 二次函数有以下形式:
  1. 一般式: y = a x 2 + b x + c {\displaystyle y=ax^{2}+bx+c}
  2. 顶点式:若二次函数顶点为 ( h , k ) {\displaystyle (h,k)} ,二次项系数为a,则:
    y = a ( x h ) 2 + k {\displaystyle y=a(x-h)^{2}+k}
  3. 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)} ,二次项系数为a,则:
    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})} 和点 B ( x 2 , y 2 ) {\displaystyle B(x_{2},y_{2})} ,则点A,B间距离为:
    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})} ,直线 l : A x + B y + C = 0 {\displaystyle l:Ax+By+C=0} ,则点P到直线l距离为:
    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}} ,则两直线间距离为:
    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}
    直线l1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}}}}
    直线l1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:

  1. 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 }}
  2. 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})} 和点 B ( x 2 , y 2 ) {\displaystyle B(x_{2},y_{2})} ,则线段AB中点坐标为
    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})} ,点 B ( x 2 , y 2 ) {\displaystyle B(x_{2},y_{2})} ,点 C ( x 3 , y 3 ) {\displaystyle C(x_{3},y_{3})} ,则三角形重心坐标为:
    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}})}
    注:其它几心过于复杂,宜自行推导
  • O {\displaystyle \odot O} 圆心为 ( a , b ) {\displaystyle (a,b)} ,半径为r,则圆的方程为:
    ( 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} 与圆 ( x a ) 2 + ( y b ) 2 = r 2 {\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}} 之间的位置关系判断:
    令圆心与直线间距离为d,两方程联立后判别式为 Δ {\displaystyle \Delta } ,则
    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}}}}}
  1. d > r {\displaystyle d>r\Leftrightarrow } 相离 Δ < 0 {\displaystyle \Leftrightarrow \Delta <0}
  2. d = r {\displaystyle d=r\Leftrightarrow } 相切 Δ = 0 {\displaystyle \Leftrightarrow \Delta =0}
  3. d < r {\displaystyle d<r\Leftrightarrow } 相交 Δ < 0 {\displaystyle \Leftrightarrow \Delta <0}

对戴芊问题的回答:函数f(x)=sin(kx)cos(jx)的周期

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}

p = k + j {\displaystyle p=k+j} q = k j {\displaystyle q=k-j} ,则有 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(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}}} 皆为 2 π {\displaystyle 2\pi } 的整数倍,即 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} } ,则 n p ,   n q {\displaystyle n\mid p,\ n\mid q} . 又由于 T {\displaystyle T} 为最小正周期, n = gcd ( | p | , | q | ) {\displaystyle n=\gcd(\left\vert p\right\vert ,\left\vert q\right\vert )} 。故 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)} 非周期函数

2) 若 ( p Q ) ( q Q ) {\displaystyle \left(p\in \mathbb {Q} \right)\cap \left(q\in \mathbb {Q} \right)} ,不妨令 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}

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 }

对戴芊问题的回答:数列通项的不动点求法

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\}} 为等差数列……①

1 a n + 1 x 1 a n x = d {\displaystyle \Leftarrow {\frac {1}{a_{n+1}-x}}-{\frac {1}{a_{n}-x}}=d} ……②

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}}} ……③

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} 。则显然①适用于形如 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}}}} 的递推关系。由于将 x {\displaystyle x} 同时带入③中的 a n + 1 ,   a n {\displaystyle a_{n+1},\ a_{n}} ,左右相等恒成立,也就证明了不动点法求解的正确性。

关于 cosx=x 的解

方程 cos x = x {\displaystyle \cos x=x} 的解为:

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}

2018年12月7日

问题一

求解: f ( x ) + f ( x ) f ( x ) = x {\displaystyle f(x)+f(x)f'(x)=x}

问题二

计算 ( 2 + i ) 1 + i {\displaystyle (2+i)^{1+i}} (以复数的代数形式表示)

问题三

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} 个字。问:至少需要经过多少次相同操作,才能使输入法的排序恢复至打出该字前?

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