Ejercicios del Álgebra de Baldor: Ejercicio 053

Ejemplo:

$Dividir \ \ \frac{3}{4} x^3y- \frac{2}{3} x^2y^2+ \frac{5}{6} xy^3- \frac{1}{2} y^4 \ \ entre \ \ \frac{5}{6}y$

$(\frac{3}{4} x^3y- \frac{2}{3} x^2y^2+ \frac{5}{6} xy^3- \frac{1}{2} y^4) \div \ \ \frac{5}{6}y$

$= \frac{ \frac{3}{4}x^3y }{ \frac{5}{6}y } - \frac{ \frac{2}{3}x^2y^2 }{\frac{5}{6}y} + \frac{ \frac{5}{6}xy^3 }{\frac{5}{6}y}- \frac{ \frac{1}{2}y^4 }{\frac{5}{6}y}$

$= \frac{ 9 }{10}x^3y^{1-1} - \frac{4}{5}x^2y^{2-1} +(1)xy^{3-1}- \frac{3}{5}y^{4-1}$

$= \frac{9}{10}x^3(1) - \frac{4}{5}x^2y^1 + xy^2 - \frac{3}{5} y^3$

$= \frac{9}{10}x^3 - \frac{4}{5}x^2y + xy^2 - \frac{3}{5} y^3 \ \ \ Solucion.$

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Ejercicio 53.

$2) \ \ \frac{1}{3}a^3 - \frac{3}{5}a^2 + \frac{1}{4}a \ \ entre \ \ -\frac{3}{5}$

$= \frac{1}{3}a^3 - \frac{3}{5}a^2 + \frac{1}{4}a \ \ \div \ \ -\frac{3}{5}$

$= \frac{ \frac{1}{3}a^3 }{- \frac{3}{5} } - \frac{ \frac{3}{5}a^2 }{- \frac{3}{5}} + \frac{ \frac{1}{4}a }{- \frac{3}{5}}}= -\frac{5}{9}a^3-(-1a^2)+(- \frac{5}{12} a)$

$= - \frac{5}{9}a^3+a^2- \frac{5}{12}a \ \ \ \ Solucion.$

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$4) \ \ \frac{2}{3}x^4y^3- \frac{1}{5}x^3y^4+ \frac{1}{4}x^2y^5-xy^6 \ \ \ entre \ \ \ - \frac{1}{5}xy^3$

$=\frac{2}{3}x^4y^3- \frac{1}{5}x^3y^4+ \frac{1}{4}x^2y^5-xy^6 \ \ \ \div \ \ \ - \frac{1}{5}xy^3$

$= \frac{\frac{2}{3}x^4y^3}{- \frac{1}{5}xy^3} - \frac{\frac{1}{5}x^3y^4}{- \frac{1}{5}xy^3} + \frac{\frac{1}{4}x^2y^5}{- \frac{1}{5}xy^3}-\frac{xy^6 }{- \frac{1}{5}xy^3}$

$=- \frac{10}{3}x^{4-1}y^{3-3}-(-1x^{3-1}y^4-3)+(- \frac{5}{4}x^{2-1}y^{}5-3 )-(-5x^{1-1}y^{6-3})$

$= - \frac{10}{3}x^3y^0+x^2y^1- \frac{5}{4}x^1y^2+5x^0y^3$

$= - \frac{10}{3}x^3+x^2y- \frac{5}{4}xy^2+5y^3 \ \ \ \ Solucion.$

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$6) \ \ \ \frac{1}{3}a^m+ \frac{1}{4}a^{m-1} \ \ \ entre \ \ \ \frac{1}{2}a$

$=\frac{1}{3}a^m+ \frac{1}{4}a^{m-1} \ \ \ \div \ \ \ \frac{1}{2}a = \frac{\frac{1}{3}a^m}{\frac{1}{2}a} + \frac{ \frac{1}{4}a^{m-1}}{\frac{1}{2}a}$

$= \frac{2}{3}a^{m-1}+ \frac{1}{2}a^{m-1-1}= \frac{2}{3}a^{m-1}+ \frac{1}{2}a^{m-2} \ \ \ \ Solucion.$

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$8) \ \ \ - \frac{3}{4}a^{n-1}x^{m+2}+ \frac{1}{8}a^nx^{m+1}- \frac{2}{3}a^{n+1}x^m \ \ \ entre \ \ - \frac{2}{5}a^3x^2$

$= - \frac{3}{4}a^{n-1}x^{m+2}+ \frac{1}{8}a^nx^{m+1}- \frac{2}{3}a^{n+1}x^m \ \ \ \div \ \ - \frac{2}{5}a^3x^2$

$= \frac{- \frac{3}{4}a^{n-1}x^{m+2}}{- \frac{2}{5}a^3x^2} + \frac{\frac{1}{8}a^nx^{m+1}}{- \frac{2}{5}a^3x^2} - \frac{ \frac{2}{3}a^{n+1}x^m}{- \frac{2}{5}a^3x^2}$

$= \frac{15}{8}a^{n-1-3}x^{m+2-2}+(- \frac{5}{16}a^{n-3} x^{m+1-2}) -(- \frac{5}{3}a^{n+1-3}x^{m-2})$

$= \frac{15}{8}a^{n-4}x^{m}- \frac{5}{16}a^{n-3} x^{m-1} + \frac{5}{3}a^{n-2}x^{m-2} \ \ \ \ Solucion.$

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Ejercicios del Álgebra de Baldor

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miércoles, 9 de junio de 2021

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