Question

Necesito pasar esto para que me promuevan de curso
(3m²-n)²
(4x+5y²)³
(2a²-4d)³
(X²+8)(x²-12)
(Y+20)(y-5)
(M-10)(x-6)
(4x²-5y)(4x²+5y)

Answer 1

▪A tomar en cuenta:

° Diferencia de cuadrados:
$\boxed{ {x}^{2} - {y}^{2} = (x + y)(x - y)}$

° Trinomio cuadrado perfecto:
$\boxed{{x}^{2} {^{ + } _{ - }} 2xy + {y}^{2} = {(x{^{ + } _{ - }}y)}^{2} }$

° Productos notables:
$\boxed{{(x{^{ + } _{ - }}y)}^{3} = ( {x}^{3} {^{ + } _{ - }}3 {x}^{2} y + 3x {y}^{2} {^{ + } _{ - }} {y}^{3} )}$

° Factor común:
$\boxed{ax+ ay = a(x + y)}$

▪Soluciones:
*en cajas.

a.
${(3 {m}^{2} - n)}^{2} \\ \\ \boxed{9 {m}^{4} - 6 {m}^{2} n + {n}^{2} }$


b.
${(4x + 5 {y}^{2} )}^{3} \\ \\ {(4x)}^{3} + 3 {(4x)}^{2}(5 {y}^{2} ) + 3(4x) {(5 {y}^{2}) }^{2} + {(5 {y}^{2}) }^{3} \\ \\ \boxed{64 {x}^{3} + 240 { x }^{2} {y}^{2} + 300x {y}^{4} + 125 {y}^{6} }$


c.
${(2 {a}^{2} - 4d)}^{3} = \\ \\ {(2 {a}^{2} )}^{3} - 3 {(2 {a}^{2}) }^{2} (4d) + 3(2 {a}^{2} ){(4d)}^{2} - {(4d)}^{3} \\ \\ \boxed{2 {a}^{6} - 48 {a}^{4} d + 96 {a}^{2} {d}^{2} - 64 {d}^{3} }$


d.
$( {x}^{2} + 8)( {x}^{2} - 12) = \\ \\ {x}^{4} - 12 {x}^{2} + 8 {x}^{2} - 96 = \\ \\ \boxed{ {x}^{4} - 4 {x}^{2} - 96}$


e.
$(y + 20)(y - 5) = \\ \\ {y}^{2} - 5y + 20y - 100 = \\ \\ \boxed{{y}^{2} + 15y - 100}$


f.
$(m - 10)(x - 6) = \\ \\ mx - 6m - 10x + 60 = \\ \\ 60 - 6m + mx - 10x = \\ \\ 6(10 - m) + x(m - 10) = \\ \\ 6(10 - m) - x( 10 - m) = \\ \\ \boxed{(10 - m)(6 - x)}$


g.
$(4 {x}^{2} - 5y)(4 {x}^{2} + 5y) = \\ \\ {(4 {x}^{2} ) }^{2} - {(5y)}^{2} = \\ \\ \boxed{ 16 {x}^{4} - 25 {y}^{4} }$