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CAPITULO V

División
Hallar el valor numérico
Ejercicio 60
Hallar el volor numérico de las expresiones siguientes para a=1,b=2,c= 1 2
  1. a 2 2ab+ b 2 = (1 ) 2 2(1 ) ( 2 ) + ( 2 ) 2 =1+4+4 =9
  2. 3 a 3 4 a 2 b+3a b 2 b 3 =3 (1 ) 3 4 ( 1 ) 2 ( 2 ) +3(1 ) ( 2 ) 2 ( 2 ) 3 =3(1 ) 8128 =328 =31
  3. a 4 3 a 3 +2ac3bc = (1 ) 4 3 (1 ) 3 + 2 (1 ) ( 1 2 ) 3(2 ) ( 1 2 ) =1+3+1+3 =8
  4. a 5 8 a 4 c+16 a 3 c 2 20 a 2 c 3 +40a c 4 c 5 = (1 ) 5 (1 ) 4 ( 1 2 ) +16 (1 ) 3 ( 1 2 ) 2 20 (1 ) 2 ( 1 2 ) 3 +40(1 ) ( 1 2 ) 4 ( 1 2 ) 5 =1+4( 1 4 ) ( 1 ) ( 1 ) + 1 32 =1+ 4 4 + 5 2 5 2 + 1 32 = 32+1 32 = 31 32
  5. (ab ) 2 + (bc ) 2 (ac ) 2 = (12 ) 2 + (2[ 1 2 ] ) 2 (1[ 1 2 ] ) 2 = (3 ) 2 + (2+ 1 2 ) 2 (1+ 1 2 ) 2 =9+ ( 4+1 2 ) 2 ( 2+1 2 ) 2 =9+ ( 5 2 ) 2 ( 1 2 ) 2 =9+ 25 4 1 4 = 36+251 4 = 4 =15
  6. (b+a ) 3 (bc ) 3 (ac ) 3 = (2+[1 ] ) 3 (2[ 1 2 ] ) 3 (1[ 1 2 ] ) 3 = (21 ) 3 (2+ 1 2 ) 3 (1+ 1 2 ) 3 = ( 1 ) 3 ( 4+1 2 ) 3 ( 2+1 2 ) 3 =1 ( 5 2 ) 3 ( 1 2 ) 3 =1 125 8 + 1 8 = 8125+1 8 = = 29 2 =14 1 2
  7. ab c + ac b bc a = (1 ) ( 2 ) 1 2 + (1 ) ( 1 2 ) 2 (2 ) ( 1 2 ) (1 ) =4+ 1 4 1 = 16+14 4 = 13 3 =3 1 4
  8. (a+b+c ) 2 (abc ) 2 +c = ([1 ] +2+[ 1 2 ] ) 2 ([1 ] 2[ 1 2 ] ) 2 +( 1 2 ) = (1 1 2 ) 2 (3+ 1 2 ) 2 1 2 = ( 21 2 ) 2 ( 6+1 2 ) 2 1 2 = ( 1 2 ) 2 ( 5 2 ) 2 1 2 = 1 4 25 4 1 2 = 1252 4 = = 13 2 =6 1 2
  9. 3(2a+b ) 4a(b+c ) 2c(ab ) =3[2(1 ) +2 ] 4(1 ) [2+( 1 2 ) ] 2 ( 1 2 ) [(1 ) 2 ] =+4( 41 2 ) +(3 ) =( 3 2 ) 3 =63 =3
Hallar el valor numérico de las expresiones siguientes para a=2,b= 1 3 ,x=2,y=1,m=3,n= 1 2
  1. x 4 8 x 2 y 2 + 3x y 2 2 y 3 = (2 ) 4 8 (2 ) 2 (1 ) 2 + 3( 2 ) (1 ) 2 2 (1 ) 3 =2+23+1 =2
  2. (ax ) 2 + (xy ) 2 +( x 2 y 2 ) (m+xn ) = (2[2 ] ) 2 + (2[1 ] ) 2 +( [2 ] 2 [1 ] 2 ) (3+[2 ] 1 2 ) = ( 4 ) 2 + (1 ) 2 +(41 ) ( 641 2 ) =16+1+3( 1 2 ) =17+ 3 2 = 34+3 2 = 37 2 =18 1 2
  3. (xy ) +( x 2 + y 2 ) (xym ) +3b(x+y+n ) =[2(1 ) ] +[ (2 ) 2 + (1 ) 2 ] [2(1 ) 3 ] + 3 ( 1 3 ) (21+ 1 2 ) =[2+1 ] +[4+1 ] [5+1 ] +( 42+1 2 ) =1+5(4 ) +( 5 2 ) =120 5 2 = 2405 2 = 43 2 =21 1 2
  4. (3x2y ) (2n4m ) +4 x 2 y 2 xy 2 =(3(2 ) 2(1 ) ) ( 2 ( 1 2 ) 4( 3 ) ) +4 (2 ) 2 (1 ) 2 2(1 ) 2 =(6+2 ) (112 ) +4( 4 ) ( 1 ) 2+1 2 =4(11 ) +16+ 1 2 =44+16+ 1 2 = 88+32+1 2 = 121 2 =60 1 2
  5. 4x 3y x 2 2+ y 3 +( 1 n 1 b ) x+ x 4 m = 4(2 ) 3(1 ) (2 ) 3 2+ (1 ) 3 +( 1 1 2 1 1 3 ) (2 ) + (2 ) 4 3 = 8 3 + 8 21 +(23 ) (2 ) +163 = 8 3 +8+(1 ) (2 ) +13 = 8 3 +21+2 = 8+63+6 3 = 77 3 =25 2 3
  6. x 2 (xy+m ) (xy ) ( x 2 + y 2 n ) + (x+y ) 2 ( m 2 2n ) = (2 ) 2 [2(1 ) +3 ] [2(1 ) ] [ (2 ) 2 + (1 ) 2 1 2 ] + [2+(1 ) ] 2 [ 3 2 2 ( 1 2 ) ] =4(2+1+3 ) (2+1 ) (4+1 1 2 ) + (21 ) 2 (91 ) =4( 2 ) (1 ) ( 8+21 2 ) + (3 ) 2 ( 8 ) =8+ 9 2 +72 = 16+9+144 2 = 169 2 =84 1 2
  7. 3a x + 2y m + 3n y m n +2( x 3 y 2 +4 ) = 3(2 ) 2 + 2(1 ) 3 + 3( 1 2 ) 1 3 1 2 +2[ (2 ) 3 (1 ) 2 +4 ] =3 2 3 3 2 6+2(81+4 ) =9 2 3 3 2 +2(5 ) = 544960 6 = 127 6 =21 1 6