Respuesta:
Uso del caso de factorización "Diferencia de cuadrados".
a²-b² = (a+b)*(a-b)
Explicación paso a paso:
1)
(x-y)*(x+y)
= (x)²-(y)²
= x²-y²
2)
(1-3ax)*(1+3ax)
= (1)²-(3ax)²
= 1- 9a²x²
3)
(a-x)*(a+x)
= (a)²-(x)²
= a²-x²
4)
[tex](\frac{1}{4}m + \frac{2}{5} n)* (\frac{1}{4}m - \frac{2}{5} n)[/tex]
= [tex](\frac{1}{4} m)^{2} - (\frac{2}{5} n)^{2}\\[/tex]
=[tex]\frac{1}{16} m^{2} - \frac{4}{25} n^{2}[/tex]>
5)
(2a-1)*(2a+1)
= (2a)²-(1)²
= 4a²-1
6)
(a-b)*(a+b)
= (a)²-(b)²
= a²-b²
7)
(m+n)*(m-n)
=(m)²-(y)²
=m²-y²
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Verified answer
Respuesta:
Uso del caso de factorización "Diferencia de cuadrados".
a²-b² = (a+b)*(a-b)
Explicación paso a paso:
1)
(x-y)*(x+y)
= (x)²-(y)²
= x²-y²
2)
(1-3ax)*(1+3ax)
= (1)²-(3ax)²
= 1- 9a²x²
3)
(a-x)*(a+x)
= (a)²-(x)²
= a²-x²
4)
[tex](\frac{1}{4}m + \frac{2}{5} n)* (\frac{1}{4}m - \frac{2}{5} n)[/tex]
= [tex](\frac{1}{4} m)^{2} - (\frac{2}{5} n)^{2}\\[/tex]
=[tex]\frac{1}{16} m^{2} - \frac{4}{25} n^{2}[/tex]>
5)
(2a-1)*(2a+1)
= (2a)²-(1)²
= 4a²-1
6)
(a-b)*(a+b)
= (a)²-(b)²
= a²-b²
7)
(m+n)*(m-n)
=(m)²-(y)²
=m²-y²