1 2 {\displaystyle {\frac {1}{2}}} 1 3 {\displaystyle {\frac {1}{3}}} 1 4 {\displaystyle {\frac {1}{4}}} 4 5 {\displaystyle {\frac {4}{5}}} 5 6 {\displaystyle {\frac {5}{6}}}
5 {\displaystyle {\sqrt {5}}} 6 3 {\displaystyle {\sqrt[{3}]{6}}} 7 4 {\displaystyle {\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 = a [ x 2 + b a x + c a ] = a [ x 2 + 2 ∗ b 2 a + ( b 2 a ) 2 − ( b 2 a ) 2 + c a ] {\displaystyle y=f(x)=ax^{2}+bx+c=a[x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}]=a[x^{2}+2*{\frac {b}{2a}}+({\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 ] = a [ ( x + b 2 a ) 2 − b 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+{\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 ] = 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+{\frac {b}{2a}})^{2}-({\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})^{2}]}
= a [ ( x + b 2 a + b 2 − 4 a c 4 a 2 ) ( x + b 2 a − b 2 − 4 a c 4 a 2 ) ] = a [ x + b ± b 2 − 4 a c 2 a ] {\displaystyle =a[(x+{\frac {b}{2a}}+{\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})(x+{\frac {b}{2a}}-{\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})]=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 , y ) , B ( x , y ) {\displaystyle A(x,y),B(x,y)}
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 \angle {A,B,C}} 對應邊 a ¯ , b ¯ , c ¯ {\displaystyle {\overline {a}},{\overline {b}},{\overline {c}}}
C 2 = a 2 + b 2 − 2 a b c o s C {\displaystyle C^{2}=a^{2}+b^{2}-2abcosC}
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}}}}}
△ t v = △ t 1 − ( v c ) 2 {\displaystyle \vartriangle {t_{v}}={\frac {\vartriangle {t}}{\sqrt {1-({\frac {v}{c}})^{2}}}}}
![{\displaystyle {\sqrt[{3}]{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eef1a86e652dc16d1bfaed26bee002199501bf4d)
![{\displaystyle {\sqrt[{4}]{7}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c7fc2a4722d158a05d51ae186218fb442b74905)
![{\displaystyle y=f(x)=ax^{2}+bx+c=a[x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}]=a[x^{2}+2*{\frac {b}{2a}}+({\frac {b}{2a}})^{2}-({\frac {b}{2a}})^{2}+{\frac {c}{a}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c9fc84542cfc3f14739dc0a9f4822c4e78607d7)
![{\displaystyle =a[(x+{\frac {b}{2a}})^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {4ac}{4a^{2}}}]=a[(x+{\frac {b}{2a}})^{2}-{\frac {b^{2}-4ac}{4a^{2}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bdffd6a22fad26ebbc57fcf4e944f611c0730fa)
![{\displaystyle =a[(x+{\frac {b}{2a}})^{2}-({\sqrt {\frac {b^{2}-4ac}{4a^{2}}}})^{2}]=a[(x+{\frac {b}{2a}})^{2}-({\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a00644c5cca062b6e20cd2b5cfeb1ac3a264af2)
![{\displaystyle =a[(x+{\frac {b}{2a}}+{\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})(x+{\frac {b}{2a}}-{\frac {\sqrt {b^{2}-4ac}}{4a^{2}}})]=a[x+{\frac {b\pm {\sqrt {b^{2}-4ac}}}{2a}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca3a7e4e28825d8d48490424cd804c356df4fd1e)
![{\displaystyle =a[x-{\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50ede7d1fa49902c93ca91add02de6c5e53e862d)
角
對應邊
