Respuesta:
Armamos un sistema de ecuaciones
1)X+Y=6
2) XY = 7
Despejando X en el 1)
\begin{gathered}X+Y=6 \\ \\ X=6-Y\end{gathered}
X+Y=6
X=6−Y
Sustituyendo el valor de X en la 2)
\begin{gathered}XY=7 \\ \\ (6-Y)Y=7 \\ \\ 6Y-Y^2=7 \\ \\ Y^2-6Y+7=0\end{gathered}
XY=7
(6−Y)Y=7
6Y−Y
2
=7
Y
−6Y+7=0
Aplicando formula ecuación cuadrática:
a = 1
b = -6
c = 7
\begin{gathered}X= \dfrac{-b\ñ \sqrt{b^2-4ac} }{2a} \\ \\ X=\dfrac{-(-6)\ñ \sqrt{(-6)^2-4(1)(7)} }{2(1)} \\ \\ X=\dfrac{6\ñ\sqrt{36-28} }{2} \\ \\ X=\dfrac{6\ñ \sqrt{8} }{2} \\ \\ X_1=\dfrac{6+ \sqrt{2\cdot 2^2} }{2}= \dfrac{6+2 \sqrt{2} }{2} =3+ \sqrt{2} \\ \\ X_2=\dfrac{6- \sqrt{2\cdot 2^2} }{2}= \dfrac{6-2 \sqrt{2} }{2} =3- \sqrt{2}\end{gathered}
X=
2a
−b\ñ
b
−4ac
2(1)
−(−6)\ñ
(−6)
−4(1)(7)
6\ñ
36−28
8
X
1
=
6+
2⋅2
6+2
=3+
6−
6−2
=3−
Tenemos el valor de Y (tomando el positivo), cualquiera de los dos funciona, lo reemplazamos en el de X
\begin{gathered}X=6-Y \\ \\ X=6-(3+ \sqrt{2}) \\ \\ X=6-3- \sqrt{2} \\ \\ X=3- \sqrt{2} \end{gathered}
X=6−(3+
)
X=6−3−
X=3−
X = 3 -√2
Y = 3+√2
Ahora aplicamos para : X³ + Y³ (a+b)³=a³+3a²b+2ab²+b³
\begin{gathered}Y=(3+ \sqrt{2})^3 \\ \\Y= 3^3+3(3)^2( \sqrt{2}) +3(3)(\sqrt{2})^2+(\sqrt{2})^3 \\ \\ Y=27+27\sqrt{2}+18+( \sqrt{2} )^3 \\ \\ Y=45+27\sqrt{2}+(\sqrt{2})^3\end{gathered}
Y=(3+
3
Y=3
+3(3)
(
)+3(3)(
+(
Y=27+27
+18+(
Y=45+27
\begin{gathered}X=(3-\sqrt{2})^3 \\ \\ X=3^3-3(3)^2(\sqrt{2})+3(3)(\sqrt{2})^2-(\sqrt{2})^3 \\ \\ X=27-27\sqrt{2}+18-(\sqrt{2})^3 \\ \\ X=45-27\sqrt{2}-(\sqrt{2})^3\end{gathered}
X=(3−
X=3
−3(3)
−(
X=27−27
+18−(
X=45−27
Teniendo los valores sumamos:
X + Y
\begin{gathered}45-27\sqrt{2}-(\sqrt{2})^3+45+27\sqrt{2}+(\sqrt{2})^3 \\ \\ 45+45-27\sqrt{2}+27\sqrt{2}-(\sqrt{2})^3+(\sqrt{2})^3 \\ \\ =90+0+0 \\ \\ =\boxed{\bf 90}\to Solucion\end{gathered}
45−27
+45+27
45+45−27
+27
=90+0+0
90
→Solucion
Explicación paso a paso:
dame coronita pliss
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Respuesta:
Armamos un sistema de ecuaciones
1)X+Y=6
2) XY = 7
Despejando X en el 1)
\begin{gathered}X+Y=6 \\ \\ X=6-Y\end{gathered}
X+Y=6
X=6−Y
Sustituyendo el valor de X en la 2)
\begin{gathered}XY=7 \\ \\ (6-Y)Y=7 \\ \\ 6Y-Y^2=7 \\ \\ Y^2-6Y+7=0\end{gathered}
XY=7
(6−Y)Y=7
6Y−Y
2
=7
Y
2
−6Y+7=0
Aplicando formula ecuación cuadrática:
a = 1
b = -6
c = 7
\begin{gathered}X= \dfrac{-b\ñ \sqrt{b^2-4ac} }{2a} \\ \\ X=\dfrac{-(-6)\ñ \sqrt{(-6)^2-4(1)(7)} }{2(1)} \\ \\ X=\dfrac{6\ñ\sqrt{36-28} }{2} \\ \\ X=\dfrac{6\ñ \sqrt{8} }{2} \\ \\ X_1=\dfrac{6+ \sqrt{2\cdot 2^2} }{2}= \dfrac{6+2 \sqrt{2} }{2} =3+ \sqrt{2} \\ \\ X_2=\dfrac{6- \sqrt{2\cdot 2^2} }{2}= \dfrac{6-2 \sqrt{2} }{2} =3- \sqrt{2}\end{gathered}
X=
2a
−b\ñ
b
2
−4ac
X=
2(1)
−(−6)\ñ
(−6)
2
−4(1)(7)
X=
2
6\ñ
36−28
X=
2
6\ñ
8
X
1
=
2
6+
2⋅2
2
=
2
6+2
2
=3+
2
X
2
=
2
6−
2⋅2
2
=
2
6−2
2
=3−
2
Tenemos el valor de Y (tomando el positivo), cualquiera de los dos funciona, lo reemplazamos en el de X
\begin{gathered}X=6-Y \\ \\ X=6-(3+ \sqrt{2}) \\ \\ X=6-3- \sqrt{2} \\ \\ X=3- \sqrt{2} \end{gathered}
X=6−Y
X=6−(3+
2
)
X=6−3−
2
X=3−
2
X = 3 -√2
Y = 3+√2
Ahora aplicamos para : X³ + Y³ (a+b)³=a³+3a²b+2ab²+b³
\begin{gathered}Y=(3+ \sqrt{2})^3 \\ \\Y= 3^3+3(3)^2( \sqrt{2}) +3(3)(\sqrt{2})^2+(\sqrt{2})^3 \\ \\ Y=27+27\sqrt{2}+18+( \sqrt{2} )^3 \\ \\ Y=45+27\sqrt{2}+(\sqrt{2})^3\end{gathered}
Y=(3+
2
)
3
Y=3
3
+3(3)
2
(
2
)+3(3)(
2
)
2
+(
2
)
3
Y=27+27
2
+18+(
2
)
3
Y=45+27
2
+(
2
)
3
\begin{gathered}X=(3-\sqrt{2})^3 \\ \\ X=3^3-3(3)^2(\sqrt{2})+3(3)(\sqrt{2})^2-(\sqrt{2})^3 \\ \\ X=27-27\sqrt{2}+18-(\sqrt{2})^3 \\ \\ X=45-27\sqrt{2}-(\sqrt{2})^3\end{gathered}
X=(3−
2
)
3
X=3
3
−3(3)
2
(
2
)+3(3)(
2
)
2
−(
2
)
3
X=27−27
2
+18−(
2
)
3
X=45−27
2
−(
2
)
3
Teniendo los valores sumamos:
X + Y
\begin{gathered}45-27\sqrt{2}-(\sqrt{2})^3+45+27\sqrt{2}+(\sqrt{2})^3 \\ \\ 45+45-27\sqrt{2}+27\sqrt{2}-(\sqrt{2})^3+(\sqrt{2})^3 \\ \\ =90+0+0 \\ \\ =\boxed{\bf 90}\to Solucion\end{gathered}
45−27
2
−(
2
)
3
+45+27
2
+(
2
)
3
45+45−27
2
+27
2
−(
2
)
3
+(
2
)
3
=90+0+0
=
90
→Solucion
Explicación paso a paso:
dame coronita pliss