Matemáticas, pregunta formulada por jochy201, hace 1 año

hola me puede ayudar por favor:

(3x^3+6y^2) (2x-5y^3)=

(-4x^2y+2z^2) (3xy^2-5z^3)=

(5ex^3-6dy^2) (-ex-5dy^3)=

(-8x^3+y^2) (-9x-6y^3)=

(4x^5+7y^3) (3x^4-7y^2)=

(ax^2+8y^2) (7ax-9y^4)=

(5bx^3+6y^2) (2bx-5y^3)=

(2gy^3+8r^2) (gy-7r^3)=

(9xw^3+5y^2) (2wx-5y^3)=

(3p^2q^3+6ef^2) (2pq-5e^4f)=

explica el procedimiento por favor

Respuestas a la pregunta

Contestado por JMC123
0
▪A tomar en cuenta

° Factor común:
 \boxed{ab + ac = a(b + c)}

° Factor común por agrupación de términos:
 \boxed{ax+ ay+ bx + by} = \\ \\ (ax + ay) + (bx + by) = \\ \\ a(x + y) + b(x + y) = \\ \\ \boxed{(x + y)(a + b)}

° Propiedad de potenciación:
 \boxed{ {a}^{b} \: ^{.} \: {a}^{c} = {a}^{b + c} }

° Propiedad conmutativa:
 \boxed{(a + b)(x + y) = ax + ay + bx + by}

° Ley de signos:
 + \: ^{.} \: + = + \\ + \: ^{.} \: - = - \\ - \: ^{.} \: - = + \\ - \: ^{.} \: + = -

▪Soluciones:
° Aplicando todo lo anterior obtenemos:

a). \\ (3 {x}^{3} + 6 {y}^{2} )(2x - 5 {y}^{3} ) = \\ \\ \boxed{6 {x}^{4} - 15 {x}^{3} {y}^{3} + 12x {y}^{2} - 30 {y}^{5} } = \\ \\ 6 {x}^{4} + 12x {y}^{2} - 15 {x}^{3} {y}^{3} - 30 {y}^{5} = \\ \\ 6x( {x}^{3} + 2 {y}^{2} ) - 15 {y}^{3} ( {x}^{3} + 2 {y}^{2} ) = \\ \\ \boxed{\boxed{( {x}^{3} + 2 {y}^{2} )(6x - 15 {y}^{3} )}}

b). \\ (- 4 {x}^{2} y + 2 {z}^{2} )(3x {y}^{2} - 5 {z}^{3} ) \\ \\ \boxed{- 12 {x}^{3} {y}^{3} + 20 {x}^{2} y {z}^{3} + 6x {y}^{2} {z}^{2} - 10 {z}^{5} }= \\ \\ 6x {y}^{2} {z}^{2} - 12 {x}^{3} {y}^{3} - 10 {z}^{5} + 20 {x}^{2} y {z}^{3} = \\ \\ 6x {y}^{2} ( {z}^{2} - 2 {x}^{2} y) - 10 {z}^{3} ( {z}^{2} - 2 {x}^{2} y) = \\ \\ \boxed{\boxed{( {z}^{2} - 2 {x}^{2} y)(6x {y}^{2} - 10 {z}^{3} )}}

c). \\ (5e {x}^{3} - 6d {y}^{2} )( - ex - 5d {y}^{3} ) = \\ \\ \boxed{- 5 {e}^{2} {x}^{4} - 25de {x}^{3} {y}^{3} + 6de {y}^{2} + 30 {d}^{2} {y}^{5} } = \\ \\ 6de {y}^{2} x + 30 {d}^{2} {y}^{5} - 5 {e}^{2} {x}^{4} - 25de {x}^{3} {y}^{3} = \\ \\ 6d {y}^{2} (e x+ 5d {y}^{3} ) - 5e {x}^{3} e(ex + 5d {y}^{3} ) = \\ \\ \boxed{\boxed{(ex + 5d {y}^{3} )(6d {y}^{2} - 5e {x}^{3} )}}

d). \\ ( - 8 {x}^{3} + {y}^{2} )( - 9x - 6 {y}^{3} ) = \\ \\ \boxed{72 {x}^{4} + 48 {x}^{3} {y}^{3} - 9x {y}^{2} - 6 {y}^{5} }= \\ \\ - 6 {y}^{5} + 48 {x}^{3} {y}^{3} - 9x {y}^{2} + 72 {x}^{4} = \\ \\ - 6 {y}^{3} ( {y}^{2} - 8 {x}^{3} ) - 9x( {y}^{2} - 8 {x}^{3} ) = \\ \\ \boxed{\boxed{( {y}^{2} - 8 {x}^{3} )( - 6 {y}^{3} - 9x)}}

e). \\ (4 {x}^{5} + 7 {y}^{3} )(3 {x}^{4} - 7 {y}^{2} ) = \\ \\ \boxed{\boxed{12 {x}^{9} - 28 {x}^{5} {y}^{2} + 21 {x}^{4} {y}^{3} - 49 {y}^{5} }}

f). \\ (a {x}^{2} + 8 {y}^{2} )(7ax - 9 {y}^{4} ) = \\ \\ 7 {a}^{2} {x}^{3} - 9a {x}^{2} {y}^{4} + 56ax {y}^{2} - 72 {y}^{6} = \\ \\ \boxed{\boxed{7 {a}^{2} {x}^{3} + 56ax {y}^{2} - 9a {x}^{2} {y}^{4} - 72 {y}^{6} }}

g). \\ (5b {x}^{3} + 6 {y}^{2} )(2bx - 5 {y}^{3} ) = \\ \\ \boxed{\boxed{10 {b}^{2} {x}^{4} - 25b {x}^{3} {y}^{3} + 12bx {y}^{2} - 30 {y}^{5} }}

h). \\ (2g {y}^{3} + 8 {r}^{2} )(gy - 7 {r}^{3} ) = \\ \\\boxed{ \boxed{2 {g}^{2} {y}^{4} - 14g {r}^{3} {y}^{3} + 8g {r}^{2} y - 56 {r}^{5} }}

i). \\ (9x {w}^{3} + 5 {y}^{2} )(2wx - 5 {y}^{3} ) = \\ \\ \boxed{\boxed{18{w}^{4} {x}^{2} - 45 {w}^{3} x{y}^{3} + 10wx {y}^{2} - 25 {y}^{5} }}

j). \\ (3 {p}^{2} {q}^{3} + 6 {ef}^{2} )(2pq - 5 {e}^{4} f) = \\ \\\boxed{ \boxed{6 {p}^{3} {q}^{4} - 15 {e}^{4} f {p}^{2} {q}^{4} + 12e {f}^{2} pq - 30 {e}^{5} {f}^{3} }}

° Salu2, JMC.
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