Respuesta:
[tex]9[/tex]
Explicación paso a paso:
[tex]x ; y :[/tex] los dos números.
El mayor promedio es la media aritmética:
[tex]M.A= \frac{x+y}{2}[/tex]
El menor promedio es la media armónica:
[tex]M.H= \frac{2xy}{x+y}[/tex]
[tex]M.A = 10[/tex] ; [tex]M.H = 8.1[/tex]
Sistema de ecuaciones:
[tex]\frac{x+y}{2} = 10, entonces: x+y = 20[/tex]
[tex]\frac{2xy}{x+y} = 8.1, entonces: \frac{2vx}{20} =8.1[/tex]
Por el método de sustitución:
[tex]x+y = 20[/tex] [tex]ecuac.1[/tex]
[tex]xy = 81[/tex] [tex]ecuac.2[/tex]
Despejamos " y " en la ecuac.1
[tex]y = 20-x[/tex]
Sustituimos " y " en la ecuac.2
[tex]x ( 20-x ) = 81[/tex]
[tex]20x -x^{2} = 81[/tex]
[tex]-x^{2} +20x-81=0[/tex]
[tex]a = -1 ; b = 20 ; c = -81[/tex]
Por la fórmula general:
[tex]x = \frac{-b\frac{+}{} \sqrt{b^{2} -4ac} }{2a} = \frac{-20\frac{+}{} \sqrt{(20)^{2} -4(-1)(-81)} }{2(-1)} =\frac{-20\frac{+}{} \sqrt{400-324} }{-2} = \frac{-20\frac{+}{} \sqrt{76} }{-2}[/tex]
[tex]x_{1} = \frac{-20+\sqrt{76} }{-2}=\frac{-20+2\sqrt{19} }{-2} =10-\sqrt{19}[/tex]
[tex]x_{2} = \frac{-20-2\sqrt{19} }{-2} = 10+\sqrt{19}[/tex]
[tex]y_{1} = 20 -10+\sqrt{19} = 10+\sqrt{19}[/tex]
[tex]y_{2} = 20-10-\sqrt{19} =10 -\sqrt{19}[/tex]
La media geométrica:
[tex]M.G = \sqrt[n]{x_{1} *y_{1} }[/tex]
[tex]M.G = \sqrt[2]{(10-\sqrt{19})*(10+\sqrt{19} ) }[/tex]
[tex]M.G = \sqrt{(10)^{2} -(\sqrt{19} )^{2}[/tex]
[tex]M.G = \sqrt{100-19}[/tex]
[tex]M.G = \sqrt{81}[/tex]
[tex]Luego: M.G = 9[/tex]
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Respuesta:
[tex]9[/tex]
Explicación paso a paso:
[tex]x ; y :[/tex] los dos números.
El mayor promedio es la media aritmética:
[tex]M.A= \frac{x+y}{2}[/tex]
El menor promedio es la media armónica:
[tex]M.H= \frac{2xy}{x+y}[/tex]
[tex]M.A = 10[/tex] ; [tex]M.H = 8.1[/tex]
Sistema de ecuaciones:
[tex]\frac{x+y}{2} = 10, entonces: x+y = 20[/tex]
[tex]\frac{2xy}{x+y} = 8.1, entonces: \frac{2vx}{20} =8.1[/tex]
Por el método de sustitución:
[tex]x+y = 20[/tex] [tex]ecuac.1[/tex]
[tex]xy = 81[/tex] [tex]ecuac.2[/tex]
Despejamos " y " en la ecuac.1
[tex]y = 20-x[/tex]
Sustituimos " y " en la ecuac.2
[tex]x ( 20-x ) = 81[/tex]
[tex]20x -x^{2} = 81[/tex]
[tex]-x^{2} +20x-81=0[/tex]
[tex]a = -1 ; b = 20 ; c = -81[/tex]
Por la fórmula general:
[tex]x = \frac{-b\frac{+}{} \sqrt{b^{2} -4ac} }{2a} = \frac{-20\frac{+}{} \sqrt{(20)^{2} -4(-1)(-81)} }{2(-1)} =\frac{-20\frac{+}{} \sqrt{400-324} }{-2} = \frac{-20\frac{+}{} \sqrt{76} }{-2}[/tex]
[tex]x_{1} = \frac{-20+\sqrt{76} }{-2}=\frac{-20+2\sqrt{19} }{-2} =10-\sqrt{19}[/tex]
[tex]x_{2} = \frac{-20-2\sqrt{19} }{-2} = 10+\sqrt{19}[/tex]
[tex]y_{1} = 20 -10+\sqrt{19} = 10+\sqrt{19}[/tex]
[tex]y_{2} = 20-10-\sqrt{19} =10 -\sqrt{19}[/tex]
La media geométrica:
[tex]M.G = \sqrt[n]{x_{1} *y_{1} }[/tex]
[tex]M.G = \sqrt[2]{(10-\sqrt{19})*(10+\sqrt{19} ) }[/tex]
[tex]M.G = \sqrt{(10)^{2} -(\sqrt{19} )^{2}[/tex]
[tex]M.G = \sqrt{100-19}[/tex]
[tex]M.G = \sqrt{81}[/tex]
[tex]Luego: M.G = 9[/tex]