Matemáticas, pregunta formulada por nayelhym, hace 1 año

desarrolla como producto o cociente notable
a). (-3xy^2-1)^2
b). (ab^3c^2+5ab)^2
c). (5m^2n^3+2)^3
d). x^5-y^5/x-y
e). 8-27m^6/2-3m^2
f). 125y^12+1/5y^4+1

Respuestas a la pregunta

Contestado por Piscis04
4
(a \pm b)^2= a^2\pm 2ab+b^2 \\  \\ a) (-3xy^2-1)^2= (-3xy^2)^2+2(-3xy^2)(-1)+(-1)^2\\  \\  (-3xy^2-1)^2= (-3xy^2)^2+2(3xy^2)+(-1)^2\\  \\  \boxed{(-3xy^2-1)^2= 9x^2y^4+6xy^2+1} \\  \\  \\ b) (ab^3c^2+5ab)^2=(ab^3c^2)^2+2*(ab^3c^2)*(5ab)+(5ab)^2 \\  \\  \boxed{(ab^3c^2+5ab)^2=a^2b^6c^4+10a^2b^4c^2+25a^2b^2} \\  \\  \\

c) (5m^2n^3+2)^3= (5m^2n^3)^3+3(5m^2n^3)^2(2)+3(5m^2n^3)(2)^2+(2)^3  \\  \\ (5m^2n^3+2)^3= (125m^6n^9)+3(25m^4n^6)(2)+3(5m^2n^3)(4)+(8)  \\  \\ (5m^2n^3+2)^3= 125m^6n^9+6(25m^4n^6)+12(5m^2n^3)+8  \\  \\  \boxed{(5m^2n^3+2)^3= 125m^6n^9+150m^4n^6+60m^2n^3+8}  \\  \\  \\ d) \frac{x^5-y^5}{x-y}=  \frac{(x-y)*(y^4+x*y^3+x^2*y^2+x^3*y+x^4)}{x-y}\ simplificamos  \\  \\    \boxed{ \frac{x^5-y^5}{x-y}=(y^4+xy^3+x^2y^2+x^3y+x^4)}


e)  \frac{8-27m^6}{2-3m^2} =  \frac{(2-3m^2)(9*m^4+6*m^2+4)}{(2-3m^2)} \ simplificamos  \\  \\  \boxed{\frac{8-27m^6}{2-3m^2} =  (9*m^4+6*m^2+4)} \\  \\  \\ f)  \frac{125y^{12}+1 }{5y^4+1}=   \frac{(5*y^4+1)*(25*y^8-5*y^4+1)}{5y^4+1}  \\  \\ \frac{125y^{12}+1 }{5y^4+1}=   \frac{(1)*(25*y^8-5*y^4+1)}{1}  \ simplificamos \\  \\  \boxed{\frac{125y^{12}+1 }{5y^4+1}=  25y^8-5y^4+1}

Espero que te sirva, salu2!!!!
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