Ejercicio 20 – Solucionario Baldor

Comparte esto 👍👍

DESCARGA

CAPITULO II

Resta

Ejercicio 20

Se identifican los términos de la resta es decir el minuendo y el sustraendo
Restamos del minuendo el sustraendo cambiando su signo cambiado
Se reduce las cantiades semejantes

De:

-8 restar 5
-8-5=-13
-7 restar 4
-7-4=-11
8 restar 11
8-11=-3
-8 restar -11
-8-(-11)=-8+11=3
-1 restar -9
-1-(-9)=-1+9=8
$2 a restar 3 b$
$2 a - 3 b$
$3 b restar 2$
$3 b - 2$
$4 x restar 6 b$
$4 x - 6 b$
$- 5 a restar 6 b$
$- 5 a - 6 b$
$- 8 x restar - 3$
$- 8 x - (- 3) = - 8 x + 3$
$- 9 a^{2} restar 5 b^{2}$
$- 9 a^{2} - 5 b^{2}$
$- 7 x y restar - 5 y z$
$- 7 x y - (- 5 y z) = - 7 x y + 5 y z$
$3 a restar 4 a$
$3 a - 4 a = - a$
$11 m^{2} restar 25 m^{2}$
$11 m^{2} - 25 m^{2} = - 14 m^{2}$
$- 6 x^{2} y restar - x^{2} y$
$- 6 x^{2} y - (- x^{2} y) = - 6 x^{2} y + x^{2} y = - 5 x^{2} y$
$11 a^{3} m^{2} restar - 7 a^{3} m^{2}$
$11 a^{3} m^{2} - (- 7 a^{3} m^{2}) = 11 a^{3} m^{2} + 7 a^{3} m^{2} = 18 a^{3} m^{2}$
$- 8 a b^{2} restar - 8 a b^{2}$
$- 8 a b^{2} - (- 8 a b^{2}) = - 8 a b^{2} + 8 a b^{2} = 0$
$31 x^{2} y restar - 46 x^{2} y$
$31 x^{2} y - (- 46 x^{2} y) = 31 x^{2} y + 46 x^{2} y = 77 x^{2} y$
$- 84 a^{2} b restar - 84 a^{2} b$
$- 84 a^{2} b - (- 84 a^{2} b) = - 84 a^{2} b + 84 a^{2} b = 0$
$3 a^{x + 1} restar 5 b^{x + 2}$
$3 a^{x + 1} - 5 b^{x + 2}$
$- 8 x^{a + 2} restar 11$
$- 8 x^{a + 2} - 11$
$6 a^{n} restar - 5 a^{n}$
$6 a^{n} - (- 5 a^{n}) = 6 a^{n} + 5 a^{n} = 11 a^{n}$
$- 45 a^{x - 1} restar - 60 a^{x - 1}$
$- 45 a^{x - 1} - (- 60 a^{x - 1}) = - 45 a^{x - 1} + 60 a^{x - 1} = 15 a^{x - 1}$
$54 b^{n - 1} restar - 86 b^{n - 1}$
$54 b^{n - 1} - (- 86 b^{n - 1}) = 54 b^{n - 1} + 86 b^{n - 1} = 140 b^{n - 1}$
$- 35 m^{a} restar - 60 m^{a}$
$- 35 m^{a} - (- 60 m^{a}) = - 35 m^{a} + 60 m^{a} = 25 m^{a}$
5 restar $- \frac{1}{2}$
$5 - (- \frac{1}{2}) = 5 + \frac{1}{2} = \frac{10 + 1}{2} = \frac{11}{2}$
$- \frac{2}{3} restar - \frac{3}{4}$
$- \frac{2}{3} - (- \frac{3}{4}) = - \frac{2}{3} + \frac{3}{4} = \frac{- 8 + 9}{12} = \frac{1}{12}$
$\frac{1}{3} x^{2} restar - \frac{2}{3} x^{2}$
$\frac{1}{3} x^{2} - (- \frac{2}{3} x^{2}) = \frac{1}{3} x^{2} + \frac{2}{3} x^{2} = \frac{x^{2} + 2 x^{2}}{3} = \frac{3 x^{2}}{3} = x^{2}$
$\frac{4}{5} x^{3} y restar - \frac{5}{6} x^{3} y$
$\frac{4}{5} x^{3} y - (- \frac{5}{6} x^{3} y) = \frac{4}{5} x^{3} y + \frac{5}{6} x^{3} y = \frac{24 x^{3} y + 25 x^{3} y}{30} = \frac{49}{30} x^{3} y$
$- \frac{1}{8} a b^{2} restar - \frac{3}{4} a b^{2}$
$- \frac{1}{8} a b^{2} - (- \frac{3}{4} a b^{2}) = - \frac{1}{8} a b^{2} + \frac{3}{4} a b^{2} = \frac{- a b^{2} + 6 a b^{2}}{8} = \frac{5}{8} a b^{2}$

Restar
$3 d e - 2$
$- 2 - 3 = - 5$
$- 1 d e 7$
$7 - (- 1) = 7 + 1 = 8$
$- 5 d e - 8$
$- 8 - (- 5) = - 8 + 5 = - 3$
$- 4 d e 5$
$5 - (- 4) = 5 + 4 = 9$
$- 7 d e - 7$
$- 7 - (- 7) = - 7 + 7 = 0$
$- 5 d e 2 a$
$2 a - (- 5) = 2 a + 5$
$b d e - 3 x$
$- 3 x - b$
$5 m d e - 2 n$
$- 2 n - 5 m$
$- 6 a d e 3 b$
$3 b - (- 6 a) = 3 b + 6 a$
$- 5 a^{3} d e 8 b$
$8 b - (- 5 a^{3}) = 8 b + 5 a^{3}$
$- 9 d e - 7 a$
$- 7 a - (- 9) = - 7 a + 9$
$- 25 d e 25 a b$
$25 a b - (- 25) = 25 a b + 25$
$- a d e 3 a$
$3 a - (- a) = 3 a + a$
$- 3 b d e - 4 b$
$- 4 b - (- 3 b) = - 4 b + 3 b$
$- 11 x^{3} d e 54 x^{3}$
$54 x^{3} - (- 11 x^{3}) = 54 x^{3} + 11 x^{3} = 65 x^{3}$
$14 a^{2} b d e 78 a^{2} b$
$78 a^{2} b - 14 a^{2} b = 64 a^{2} b$
$- 43 a^{2} y d e - 54 a^{2} y$
$- 54 a^{2} y - (- 43 a^{2} y) = - 54 a^{2} y + 43 a^{2} y = - 11 a^{2} y$
$9 a b d e - a b$
$- a b - 9 a b = - 10 a b$
$- 31 x^{2} y d e - 31 x^{2} y$
$- 31 x^{2} y - (- 31 x^{2} y) = - 31 x^{2} y + 31 x^{2} y = 0$
$a^{x} d e - 3 a^{x}$
$- 3 a^{x} - a^{x} = - 4 a^{x}$
$- 7 a^{x + 1} d e 311 a^{x + 1}$
$311 a^{x + 1} - (- 7 a^{x + 1}) = 311 a^{x + 1} + 7 a^{x + 1} = 318 a^{x + 1}$
$9 m^{x} d e 105 m^{x}$
$105 m^{x} - 9 m^{x} = 94 m^{x}$
$18 a^{x - 1} d e - 31 a^{x - 1}$
$- 31 a^{x - 1} - 18 a^{x - 1} = - 49 a^{x - 1}$
$- 19 m^{a} d e - 236 m^{a}$
$- 236 m^{a} - (- 19 m^{a}) = - 236 m^{a} + 19 m^{a} = - 217 m^{a}$
$54 a^{x + 2} d e - 85 a^{x + 2}$
$- 85 a^{x + 2} - 54 a^{x + 2} = - 139 a^{x + 2}$
$- 6 a d e \frac{1}{4}$
$\frac{1}{4} - (- 6 a) = \frac{1}{4} + 6 a$
$- 5 d e - \frac{2}{3}$
$- \frac{2}{3} - (- 5) = - \frac{2}{3} + 5 = \frac{- 2 + 15}{3} = \frac{13}{3}$
$\frac{3}{8} m^{3} d e - \frac{7}{10} m^{3}$
$- \frac{7}{10} m^{3} - \frac{3}{8} m^{3} = \frac{- 28 m^{3} - 15 m^{3}}{40} = - \frac{43}{40} m^{3}$
$- \frac{11}{12} a^{2} b^{2} d e \frac{5}{6} a^{2} b^{2}$
$\frac{5}{6} a^{2} b^{2} - (- \frac{11}{12} a^{2} b^{2}) = \frac{5}{6} a^{2} b^{2} + \frac{11}{12} a^{2} b^{2} = \frac{10 a^{2} b^{2} + 11 a^{2} b^{2}}{12} = \frac{}{} a^{2} b^{2} = \frac{7}{4} a^{2} b^{2}$
$45 a^{3} b^{2} d e - \frac{1}{9} a^{3} b^{2}$
$- \frac{1}{9} a^{3} b^{2} - 45 a^{3} b^{2} = \frac{- a^{3} b^{2} - 405 a^{3} b^{2}}{9} = - \frac{406}{9} a^{3} b^{2} = - 45 \frac{1}{9} a^{3} b^{2}$