CASOS DE FACTORIZACIÓN
CASO ESPECIAL (III)
Trinomio cuadrado perfecto
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Respuestas
Respuesta:
(m−n)2+6(m−n)+9
=m2+−2mn+n2+(6)(m)+(6)(−n)+9
=m2+−2mn+n2+6m+−6n+9
Solución:
=m2−2mn+n2+6m−6n+9
(a+x)2−2(a+x)(x+y)+(x+y)2
=a2+2ax+x2+−2ax+−2ay+−2x2+−2xy+x2+2xy+y2
=a2+2ax+x2+−2ax+−2ay+−2x2+−2xy+x2+2xy+y2
=(a2)+(2ax+−2ax)+(−2ay)+(x2+−2x2+x2)+(−2xy+2xy)+(y2)
=a2+−2ay+y2
solución:
=a2−2ay+y2
(m+n)2−2(a−m)(m−n)+(a+m)2
=m2+2mn+n2+−2am+2an+2m2+−2mn+a2+2am+m2
=m2+2mn+n2+−2am+2an+2m2+−2mn+a2+2am+m2
=(a2)+(−2am+2am)+(2an)+(m2+2m2+m2)+(2mn+−2mn)+(n2)
=a2+2an+4m2+n2
solución:
=a2+2an+4m2+n2
4(1+a)2−4(1+a)(b−1)+(b−1)2
=4a2+8a+4+−4ab+4a+−4b+4+b2+−2b+1
=4a2+8a+4+−4ab+4a+−4b+4+b2+−2b+1
=(4a2)+(−4ab)+(b2)+(8a+4a)+(−4b+−2b)+(4+4+1)
=4a2+−4ab+b2+12a+−6b+9
solución:
=4a2−4ab+b2+12a−6b+9
9(x−y)2+12(x−y)(x+y)+4(x+y)2
=9x2+−18xy+9y2+(12(x−y))(x)+(12(x−y))(y)+4x2+8xy+4y2
=9x2+−18xy+9y2+12x2+−12xy+12xy+−12y2+4x2+8xy+4y2
=9x2+−18xy+9y2+12x2+−12xy+12xy+−12y2+4x2+8xy+4y2
=(9x2+12x2+4x2)+(−18xy+−12xy+12xy+8xy)+(9y2+−12y2+4y2)
=25x2+−10xy+y2
solución:
=25x2−10xy+y2
Explicación paso a paso: