Identitate nabarmen

${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\,}$

${\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\,}$

• Adibideak:

1. ${\displaystyle \left({\frac {4x}{5y}}+z\right)^{2}={\frac {16x^{2}}{25y^{2}}}+{\frac {8xz}{5y}}+z^{2}}$
2. ${\displaystyle (8x+a)^{2}=64x^{2}+16ax+a^{2}\,}$
3. ${\displaystyle (x-y)^{2}=(x-y).(x-y)=x^{2}-xy-yx+y^{2}=x^{2}-2xy+y^{2}\,}$
4. ${\displaystyle \left({\frac {3m}{4n}}-p\right)^{2}={\frac {9m^{2}}{16n^{2}}}-{\frac {6mp}{4n}}+p^{2}}$
5. ${\displaystyle (1-2x)^{2}=1-4x+4x^{2}\,}$

${\displaystyle (a+b).(a-b)=a^{2}-ab+ba-b^{2}=a^{2}-b^{2}\,}$

• Adibideak:

1. ${\displaystyle (a^{2}+b^{3}).(a^{2}-b^{3})=a^{4}-b^{6}\,}$
2. ${\displaystyle \left({\frac {a}{x}}-2\right).\left({\frac {a}{x}}+2\right)={\frac {a^{2}}{x^{2}}}-4}$

Binomio baten kuboaren bolumetria-deskonposizioa

${\displaystyle (x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\,}$

${\displaystyle (x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}\,}$

• Adibideak:

1. ${\displaystyle (m+3n)^{3}=m^{3}+9m^{2}n+27mn^{2}+27n^{3}\,}$
2. ${\displaystyle (x+2)^{3}=x^{3}+6x^{2}+12x+8\,}$
3. ${\displaystyle (b-2c)^{3}=b^{3}-6b^{2}c+12bc^{2}-8c^{3}\,}$
4. ${\displaystyle \left({\frac {x}{y}}-{\frac {a}{b}}\right)^{3}={\frac {x^{3}}{y^{3}}}-{\frac {3ax^{2}}{by^{2}}}+{\frac {3a^{2}x}{b^{2}y}}-{\frac {a^{3}}{b^{3}}}\,}$
5. ${\displaystyle (1-x)^{3}=1-3x+3x^{2}-x^{3}\,}$

${\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\,}$

• Adibideak:

1. ${\displaystyle (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2yz\,}$
2. ${\displaystyle (x-2y-3)^{2}=x^{2}+(-2y)^{2}+(-3)^{2}+2x(-2y)+2x(-3)+2(-2y)(-3)\,}$
   

${\displaystyle =x^{2}+4y^{2}+9-4xy-6x+12y\,}$

Gai komun bat duten 2 binomioen biderkadura.

${\displaystyle (x+a).(x+b)=x^{2}+(a+b)x+ab\,}$

• Adibideak:

1. ${\displaystyle (x+4)(x+3)=x^{2}+(4+3)x+4.3=x^{2}+7x+12\,}$
2. ${\displaystyle (x-2)(x-6)=x^{2}+(-2-6)x+(-2)(-6)=x^{2}-8x+12\,}$
3. ${\displaystyle (x-1)(x+5)=x^{2}+(-1+5)x+5(-1)=x^{2}+4x-5\,}$

${\displaystyle (a+b).(a^{2}-ab+b^{2})=a^{3}+b^{3}\,}$

${\displaystyle (a-b).(a^{2}+ab+b^{2})=a^{3}-b^{3}\,}$

• Adibideak:

1. ${\displaystyle (x+5)(x^{2}-5x+25)=x^{3}+5^{3}=x^{3}+125\,}$
2. ${\displaystyle (x-3).(x^{2}+3x+9)=x^{3}-3^{3}=x^{3}-27\,}$

${\displaystyle (x^{2}+x+1)(x^{2}-x+1)=x^{4}+x^{2}+1\,}$

${\displaystyle a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)\,}$

${\displaystyle a^{3}+b^{3}+c^{3}-3abc={\frac {1}{2}}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(a-c)^{2}]\,}$

${\displaystyle (a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})\,}$

${\displaystyle (a+b)^{2}-(a-b)^{2}=4ab\,}$

${\displaystyle (a+b)^{4}-(a-b)^{4}=8ab(a^{2}+b^{2})\,}$

${\displaystyle (a^{2}+b^{2})(x^{2}+y^{2})=(ax+by)^{2}+(ay-bx)^{2}\,}$

${\displaystyle (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=(ax+by+cz)^{2}+(ay-bx)^{2}+(az-cx)^{2}+(bz-cy)^{2}\,}$

• Binomioa
• Pascalen hirukia