1 2 , 1 3 , 1 4 , 4 5 , 5 6 {\displaystyle {\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {4}{5}},{\frac {5}{6}}}
5 , 6 3 , 7 4 {\displaystyle {\sqrt {5}},{\sqrt[{3}]{6}},{\sqrt[{4}]{7}}}
− b ± b 2 − 4 a c 2 a {\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
y = f ( x ) = a x 2 + b x + c {\displaystyle y=f(x)=ax^{2}+bx+c} = a [ x 2 + b a x + c a ] {\displaystyle =a[x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}]} = a [ x 2 + 2 ∗ b 2 a x + ( b 2 a ) 2 − ( b 2 a ) 2 + c a ] {\displaystyle =a[x^{2}+2*{\frac {b}{2a}}x+({\frac {b}{2a}})^{2}-({\frac {b}{2a}})^{2}+{\frac {c}{a}}]} = a [ ( x + b 2 a ) 2 − b 2 4 a 2 + 4 a c 4 a 2 ] {\displaystyle =a[(x+{\frac {b}{2a}})^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {4ac}{4a^{2}}}]} = a [ ( x + b 2 a ) 2 − b 2 − 4 a c 4 a 2 ] {\displaystyle =a[(x+{\frac {b}{2a}})^{2}-{\frac {b^{2}-4ac}{4a^{2}}}]} = a [ ( x + b 2 a ) 2 − ( b 2 − 4 a c 4 a 2 ) 2 ] {\displaystyle =a[(x+{\frac {b}{2a}})^{2}-({\sqrt {\frac {b^{2}-4ac}{4a^{2}}}})^{2}]} = a [ ( x + b 2 a ) 2 − ( b 2 − 4 a c 2 a ) 2 ] {\displaystyle =a[(x+{\frac {b}{2a}})^{2}-({\frac {\sqrt {b^{2}-4ac}}{2a}})^{2}]} = a [ ( x + b 2 a + b 2 − 4 a c 2 a ) ( x + b 2 a − b 2 − 4 a c 2 a ) ] {\displaystyle =a[(x+{\frac {b}{2a}}+{\frac {\sqrt {b^{2}-4ac}}{2a}})(x+{\frac {b}{2a}}-{\frac {\sqrt {b^{2}-4ac}}{2a}})]} = a [ x + b ± b 2 − 4 a c 2 a ] {\displaystyle =a[x+{\frac {b\pm {\sqrt {b^{2}-4ac}}}{2a}}]} = a [ x − − b ± b 2 − 4 a c 2 a ] {\displaystyle =a[x-{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}]}
A ( x 1 , y 1 ) , B ( x 2 , y 2 ) {\displaystyle A(x_{1},y_{1}),B(x_{2},y_{2})} 距離 = ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 {\displaystyle ={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}} 中點座標 = ( x 1 + x 2 2 , y 1 + y 2 2 ) {\displaystyle =({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})} A B → = ( x 2 − x 1 , y 2 − y 1 ) {\displaystyle {\overrightarrow {AB}}=(x_{2}-x_{1},y_{2}-y_{1})}
△ A B C {\displaystyle \triangle ABC} ,角 A , B , C {\displaystyle A,B,C} ,對應邊 a , b , c {\displaystyle a,b,c} c 2 = a 2 + b 2 − 2 a b cos C {\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos C}
F = G M 1 M 2 r 2 {\displaystyle F=G{\frac {M_{1}M_{2}}{r^{2}}}} a = v 2 r {\displaystyle a={\frac {v^{2}}{r}}}
m = m 0 1 − ( v c ) 2 {\displaystyle m={\frac {m_{0}}{\sqrt {1-({\frac {v}{c}})^{2}}}}} Δ t v = Δ t 1 − ( v c ) 2 {\displaystyle \Delta {t_{v}}={\frac {\Delta {t}}{\sqrt {1-({\frac {v}{c}})^{2}}}}}