Ejercicio 48

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CAPITULO IV

Multiplicación
Multiplicación de monomios
Ejercicio 48
  • Se suprimen los signos de agrupación más internos
  • Se reduce efectuando las operaciones indicadas
  • Reducimos términos semejantes
Simplificar
  1. x[3a+2(x+1 ) ] x[3a+2(x+1 ) ] = x[3a2x+2 ] = x3a+2x2 = 3x3a2
  2. (a+b ) 3[2a+b(a+2 ) ] (a+b ) 3[2a+b(a+2 ) ] = ab3[2aab+2b ] = ab6a+3ab6b = 7a+3ab7b
  3. [3x2y+(x2y ) 2(x+y ) 3(2x+1 ) ] [3x2y+(x2y ) 2(x+y ) 3(2x+1 ) ] = [3x2y+x2y2x2y6x3 ] = [4x6y3 ] = 4x+6y+3
  4. 4 x 2 {3x+5[x+x(2x ) ] } 4 x 2 {3x+5[x+x(2x ) ] } = 4 x 2 {3x+5[x+2x x 2 ] } = 4 x 2 {3x+5x+ x 2 } = 4 x 2 {54x+ x 2 } = 4 x 2 5+4x x 2 = 3 x 2 +4x5
  5. 2a{3x+2[a+3x2(a+b 2+a ¯ ) ] } 2a{3x+2[a+3x2(a+b 2+a ¯ ) ] } = 2a{3x+2[a+3x2(a+b2a ) ] } = 2a{3x+2[a+3x2(2a+b2 ) ] } = 2a{3x+2[a+3x+4a2b+4 ] } = 2a{3x+2[3a+3x2b+4 ] } = 2a{3x+6a+6x4b+8 } = 2a{6a+3x4b+8 } = 2a6a3x+4b8 = 4a+4b3x8
  6. a(x+y ) 3(xy ) +2[(x2y ) 2(xy ) ] a(x+y ) 3(xy ) +2[(x2y ) 2(xy ) ] = axy3x+3y+2[x+2y+2x+2y ] = a4x+2y+2[x+4y ] = a4x+2y+2x+8y = a2x+10y
  7. m(m+n ) 3{2m+[2m+n+2(1+n ) m+n1 ¯ ] } m(m+n ) 3{2m+[2m+n+2(1+n ) m+n1 ¯ ] } = m m n3{2m+[2m+ n 2+2nm n +1 ] } = n3{2m+[3m+2n1 ] } = n3{2m3m+2n1 } = n3{5m+2n1 } = n+15m6n+3 = 15m7n+3
  8. 2(ab ) 3(a+2b ) 4{a2b+2[a+b1+2(ab ) ] } 2(ab ) 3(a+2b ) 4{a2b+2[a+b1+2(ab ) ] } = 2a+2b3a6b4{a2b+2[a+b1+2a2b ] } = 5a4b4{a2b+2[ab1 ] } = 5a4b4{a2b+2a2b2 } = 5a4b4{3a4b2 } = 5a4b12a+16b+8 = 17a+12b+8
  9. 5(x+y ) [2xy+2{x+y3 xy1 ¯ } ] +2x 5(x+y ) [2xy+2{x+y3 xy1 ¯ } ] +2x = 5x5y[2xy+2{x+y3x+y+1 } ] +2x = 3x5y[2xy+2{2x+2y2 } ] = 3x5y[2xy4x+4y4 ] = 3x5y[2x+3y4 ] = 3x5y+2x3y+4 = x8y+4
  10. m3(m+n ) +[{(2m+n23[mn+1 ] ) +m } ] m3(m+n ) +[{(2m+n23[mn+1 ] ) +m } ] = m3m3n+[{(2m+n23m+3n3 ) +m } ] = 2m3n+[{(5m+4n5 ) +m } ] = 2m3n+[{5m4n+5+m } ] = 2m3n+[{6m4n+5 } ] = 2m3n+[6m+4n5 ] = 2m3n6m+4n5 = 8m+n5
  11. 3(x2y ) +2{4[2x3(x+y ) ] } {[(x+y ) ] } 3(x2y ) +2{4[2x3(x+y ) ] } {[(x+y ) ] } = 3x+6y+2{4[2x3x3y ] } { (x+y )} = 3x+6y+2{4[5x3y ] } xy = 4x+5y+2{20x+12y } = 4x+5y+40x+24y = 36x+29y
  12. 5{(a+b ) 3[2a+3b(a+b ) +(ab ) +2(a+b ) ] a } 5{(a+b ) 3[2a+3b(a+b ) +(ab ) +2(a+b ) ] a } = 5{ab3[2a+3ba b a b 2a+ 2b ] a } = 5{2ab3[6a+3b ] } = 5{2ab+18a9b } = 5{16a10b } = 80a50b
  13. 3{[+(a+b ) ] } 4{[(ab ) ] } 3{[+(a+b ) ] } 4{[(ab ) ] } = 3{(a+b ) } 4{ (ab )} = 3{ab } +4a+4b = 3a+3b+4a+4b = a+7b
  14. {a+b2(ab ) +3{[2a+b3(a+b1 ) ] } 3[a+2(1+a ) ] } {a+b2(ab ) +3{[2a+b3(a+b1 ) ] } 3[a+2(1+a ) ] } = {a+b2a+2b+3{[2a+b3a3b+3 ] } 3[a2+2a ] } = {a+3b+3{[a2b+3 ] } 3[2+a ] } = {a+3b+3{a+2b3 } +63a } = {4a+3b+3a+6b9+6 } = {a+9b3 } = a9b+3