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Identitate nabarmenak (baita Identitate nabariak edo Biderkadura nabarmenak ere) eragiketak egiteko askotan erabiltzen diren identitateak dira. Kalkulu aljebraikoan, zenbait adierazpen aljebraiko maiz agertzen dira, eta daukaten garrantziagatik identitate nabarmenak deritzegu.
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
{\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\,}
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\,}
(
4
x
5
y
+
z
)
2
=
16
x
2
25
y
2
+
8
x
z
5
y
+
z
2
{\displaystyle \left({\frac {4x}{5y}}+z\right)^{2}={\frac {16x^{2}}{25y^{2}}}+{\frac {8xz}{5y}}+z^{2}}
(
8
x
+
a
)
2
=
64
x
2
+
16
a
x
+
a
2
{\displaystyle (8x+a)^{2}=64x^{2}+16ax+a^{2}\,}
(
x
−
y
)
2
=
(
x
−
y
)
.
(
x
−
y
)
=
x
2
−
x
y
−
y
x
+
y
2
=
x
2
−
2
x
y
+
y
2
{\displaystyle (x-y)^{2}=(x-y).(x-y)=x^{2}-xy-yx+y^{2}=x^{2}-2xy+y^{2}\,}
(
3
m
4
n
−
p
)
2
=
9
m
2
16
n
2
−
6
m
p
4
n
+
p
2
{\displaystyle \left({\frac {3m}{4n}}-p\right)^{2}={\frac {9m^{2}}{16n^{2}}}-{\frac {6mp}{4n}}+p^{2}}
(
1
−
2
x
)
2
=
1
−
4
x
+
4
x
2
{\displaystyle (1-2x)^{2}=1-4x+4x^{2}\,}
(
a
+
b
)
.
(
a
−
b
)
=
a
2
−
a
b
+
b
a
−
b
2
=
a
2
−
b
2
{\displaystyle (a+b).(a-b)=a^{2}-ab+ba-b^{2}=a^{2}-b^{2}\,}
(
a
2
+
b
3
)
.
(
a
2
−
b
3
)
=
a
4
−
b
6
{\displaystyle (a^{2}+b^{3}).(a^{2}-b^{3})=a^{4}-b^{6}\,}
(
a
x
−
2
)
.
(
a
x
+
2
)
=
a
2
x
2
−
4
{\displaystyle \left({\frac {a}{x}}-2\right).\left({\frac {a}{x}}+2\right)={\frac {a^{2}}{x^{2}}}-4}
Binomio baten kuboaren bolumetria-deskonposizioa
(
x
+
y
)
3
=
x
3
+
3
x
2
y
+
3
x
y
2
+
y
3
{\displaystyle (x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\,}
(
x
−
y
)
3
=
x
3
−
3
x
2
y
+
3
x
y
2
−
y
3
{\displaystyle (x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}\,}
(
m
+
3
n
)
3
=
m
3
+
9
m
2
n
+
27
m
n
2
+
27
n
3
{\displaystyle (m+3n)^{3}=m^{3}+9m^{2}n+27mn^{2}+27n^{3}\,}
(
x
+
2
)
3
=
x
3
+
6
x
2
+
12
x
+
8
{\displaystyle (x+2)^{3}=x^{3}+6x^{2}+12x+8\,}
(
b
−
2
c
)
3
=
b
3
−
6
b
2
c
+
12
b
c
2
−
8
c
3
{\displaystyle (b-2c)^{3}=b^{3}-6b^{2}c+12bc^{2}-8c^{3}\,}
(
x
y
−
a
b
)
3
=
x
3
y
3
−
3
a
x
2
b
y
2
+
3
a
2
x
b
2
y
−
a
3
b
3
{\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}}}\,}
(
1
−
x
)
3
=
1
−
3
x
+
3
x
2
−
x
3
{\displaystyle (1-x)^{3}=1-3x+3x^{2}-x^{3}\,}
(
a
+
b
+
c
)
2
=
a
2
+
b
2
+
c
2
+
2
a
b
+
2
a
c
+
2
b
c
{\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\,}
(
x
+
y
+
z
)
2
=
x
2
+
y
2
+
z
2
+
2
x
y
+
2
x
z
+
2
y
z
{\displaystyle (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2yz\,}
(
x
−
2
y
−
3
)
2
=
x
2
+
(
−
2
y
)
2
+
(
−
3
)
2
+
2
x
(
−
2
y
)
+
2
x
(
−
3
)
+
2
(
−
2
y
)
(
−
3
)
{\displaystyle (x-2y-3)^{2}=x^{2}+(-2y)^{2}+(-3)^{2}+2x(-2y)+2x(-3)+2(-2y)(-3)\,}
=
x
2
+
4
y
2
+
9
−
4
x
y
−
6
x
+
12
y
{\displaystyle =x^{2}+4y^{2}+9-4xy-6x+12y\,}
Gai komun bat duten 2 binomioen biderkadura.
(
x
+
a
)
.
(
x
+
b
)
=
x
2
+
(
a
+
b
)
x
+
a
b
{\displaystyle (x+a).(x+b)=x^{2}+(a+b)x+ab\,}
(
x
+
4
)
(
x
+
3
)
=
x
2
+
(
4
+
3
)
x
+
4.3
=
x
2
+
7
x
+
12
{\displaystyle (x+4)(x+3)=x^{2}+(4+3)x+4.3=x^{2}+7x+12\,}
(
x
−
2
)
(
x
−
6
)
=
x
2
+
(
−
2
−
6
)
x
+
(
−
2
)
(
−
6
)
=
x
2
−
8
x
+
12
{\displaystyle (x-2)(x-6)=x^{2}+(-2-6)x+(-2)(-6)=x^{2}-8x+12\,}
(
x
−
1
)
(
x
+
5
)
=
x
2
+
(
−
1
+
5
)
x
+
5
(
−
1
)
=
x
2
+
4
x
−
5
{\displaystyle (x-1)(x+5)=x^{2}+(-1+5)x+5(-1)=x^{2}+4x-5\,}
(
a
+
b
)
.
(
a
2
−
a
b
+
b
2
)
=
a
3
+
b
3
{\displaystyle (a+b).(a^{2}-ab+b^{2})=a^{3}+b^{3}\,}
(
a
−
b
)
.
(
a
2
+
a
b
+
b
2
)
=
a
3
−
b
3
{\displaystyle (a-b).(a^{2}+ab+b^{2})=a^{3}-b^{3}\,}
(
x
+
5
)
(
x
2
−
5
x
+
25
)
=
x
3
+
5
3
=
x
3
+
125
{\displaystyle (x+5)(x^{2}-5x+25)=x^{3}+5^{3}=x^{3}+125\,}
(
x
−
3
)
.
(
x
2
+
3
x
+
9
)
=
x
3
−
3
3
=
x
3
−
27
{\displaystyle (x-3).(x^{2}+3x+9)=x^{3}-3^{3}=x^{3}-27\,}
(
x
2
+
x
+
1
)
(
x
2
−
x
+
1
)
=
x
4
+
x
2
+
1
{\displaystyle (x^{2}+x+1)(x^{2}-x+1)=x^{4}+x^{2}+1\,}
a
3
+
b
3
+
c
3
−
3
a
b
c
=
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
a
c
)
{\displaystyle a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)\,}
a
3
+
b
3
+
c
3
−
3
a
b
c
=
1
2
(
a
+
b
+
c
)
[
(
a
−
b
)
2
+
(
b
−
c
)
2
+
(
a
−
c
)
2
]
{\displaystyle a^{3}+b^{3}+c^{3}-3abc={\frac {1}{2}}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(a-c)^{2}]\,}
(
a
+
b
)
2
+
(
a
−
b
)
2
=
2
(
a
2
+
b
2
)
{\displaystyle (a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})\,}
(
a
+
b
)
2
−
(
a
−
b
)
2
=
4
a
b
{\displaystyle (a+b)^{2}-(a-b)^{2}=4ab\,}
(
a
+
b
)
4
−
(
a
−
b
)
4
=
8
a
b
(
a
2
+
b
2
)
{\displaystyle (a+b)^{4}-(a-b)^{4}=8ab(a^{2}+b^{2})\,}
(
a
2
+
b
2
)
(
x
2
+
y
2
)
=
(
a
x
+
b
y
)
2
+
(
a
y
−
b
x
)
2
{\displaystyle (a^{2}+b^{2})(x^{2}+y^{2})=(ax+by)^{2}+(ay-bx)^{2}\,}
(
a
2
+
b
2
+
c
2
)
(
x
2
+
y
2
+
z
2
)
=
(
a
x
+
b
y
+
c
z
)
2
+
(
a
y
−
b
x
)
2
+
(
a
z
−
c
x
)
2
+
(
b
z
−
c
y
)
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}\,}