miércoles, 10 de agosto de 2022

ecuacions polinómiques y english

ax^{2}+bx+c = 0

2ax+b = ( b^{2}+(-4)·ac )^{(1/2)}

(12) >] 1


ax^{2}+bx+c = ...

... a·(x+( 1/(2a) )·( b+(-1)·( b^{2}+(-4)·ac )^{(1/2)} ))·...

... (x+( 1/(2a) )·( b+( b^{2}+(-4)·ac )^{(1/2)} ))


| 1 | a+b | ab |

| 1 | b | ab+(-1)·ab = 0 |

x^{2}+(a+b)·x+ab = (x+a)·(x+b)

| 1 | a+b | ab |

| 1 | a | ab+(-1)·ab = 0 |

x^{2}+(a+b)·x+ab = (x+b)·(x+a)


ax^{3}+bx^{2}+cx+d = 0

x = ( y+(-1)·(1/3)·(b/a) )

y^{3}+py+q = 0

y = u+v

( u^{3}+v^{3} = (-q) & 3uv·y = (-1)·p·y )

(uv)^{3} = (-1)^{3}·(1/27)·p^{3}

(vu)^{3} = (-1)^{3}·(1/27)·p^{3}

u^{6}+(vu)^{3} = (-q)·u^{3}

(uv)^{3}+v^{6} = (-q)·v^{3}

u^{6}+(-1)^{3}·(1/27)·p^{3} = (-q)·u^{3}

v^{6}+(-1)^{3}·(1/27)·p^{3} = (-q)·v^{3}

(123) >] (13) >] 3 [< (12) >] 1


| 1 | 0 | p | q |

| 1 | s | p+h^{2} | q+ph+h^{3} = 0 |

y^{3}+py+q = (y+(-h))·( y^{2}+hy+(p+h^{2}) )

y^{3}+py+q = y^{3}+( hy^{2}+(-h)·y^{2} )+py+( h^{2}y+(-1)·h^{2}y )+(-1)·( ph+h^{3} )


p = (-1)·(1/3)·(b/a)^{2}+(c/a)

q = (2/27)·(b/a)^{3}+(-1)·(1/3)·(b/a)·(c/a)+(d/a)

ax^{3}+bx^{2}+cx+d = ...

... a·(x+( (1/3)·(b/a)+(-h) ))·( ( x+(1/3)·(b/a) )^{2}+h·( x+(1/3)·(b/a) )+(p+h^{2}) )


| 1 | a+b+c | ab+bc+ca | abc |

| 1 | a+b | ab | abc+(-1)·abc = 0 |

x^{3}+(a+b+c)·x^{2}+(ab+bc+ca)·x+abc = (x+c)·( x^{2}+(a+b)·x+ab )


x^{3}+x^{2}+x+1 = 0

y = ( x+(-1)·(1/3) )

y^{3}+py+q = 0

p = (2/3)

q = (20/27)

h = (1/3)·( (1/2)·( (-20)+432^{(1/2)} ) )^{(1/3)}+...

... (1/3)·( (1/2)·( (-20)+(-1)·432^{(1/2)} ) )^{(1/3)}

h = (-1)·(2/3)

h^{3} = ((-8)/27)

(-q)+(-p)·h = h^{3}

((-20)/27)+((-2)/3)·((-2)/3) = ((-8)/27)

x_{0} = h+(-1)·(1/3) = (-1)

x^{3}+x^{2}+x+1 = (x+1)·(x^{2}+1) = (x+1)·(x+(-i))·(x+i)


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I shut havere-kate taked a coffi wizh milk in the bar.

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and I not havere-kate taked a coffi wizh milk in the bar.


Teorema:

Si ( [Ec][ f(x) es continua a x = c ] & f(x+y) = f(x)+f(y) ) ==> [Ax][ f(x) es continua ]

Demostración:

| f(h) | = | f(c+h+(-c)) | = | f(c+h)+(-1)·f(c) | < s

| f(x+h)+(-1)·f(x) | = | f(x+h+(-x)) | = | f(h) | < s


Teorema:

Si f(x+y) = f(x)+y ==> [Ax][ f(x) es continua ]

Demostración:

| f(x+h)+(-1)·f(x) | = | f(x)+h+(-1)·f(x) | = |h| < s

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