Ответ:
[tex]\left\{\begin{array}{l}xy=\dfrac{1}{8}\\2x^2+2y^2=\dfrac{5}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\2(x^2+y^2)=\dfrac{5}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\ \ \ \Big|\cdot 2\\x^2+y^2=\dfrac{5}{16}\end{array}\right\ \ \oplus[/tex]
Умножим на 2 первое уравнение и прибавим ко второму уравнению
и применим формулу квадрата суммы: [tex](x+y)^2=x^2+y^2+2xy[/tex] .
[tex]\left\{\begin{array}{l}xy=\dfrac{1}{8}\\x^2+y^2+2xy=\dfrac{5}{16}+\dfrac{2}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\(x+y)^2=\dfrac{5}{16}+\dfrac{1}{4}\end{array}\right\\\\\\\left\{\begin{array}{l}xy=\dfrac{1}{8}\\(x+y)^2=\dfrac{9}{16}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=\pm \dfrac{3}{4}\end{array}\right[/tex]
[tex]a)\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x(-x+\dfrac{3}{4})=\dfrac{1}{8}\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}-x^2+\dfrac{3}{4}\, x-\dfrac{1}{8}=0\\y=-x+\dfrac{3}{4}\end{array}\right[/tex]
[tex]\left\{\begin{array}{l}8x^2-6x+1=0\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x_1=\dfrac{1}{4}\ ,\ x_2=\dfrac{1}{2}\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}\ x_1=\dfrac{1}{4}\ ,\ x_2=\dfrac{1}{2}\\y_1=\dfrac{1}{2}\ ,\ y_2=\dfrac{1}{4}\end{array}\right\\\\\\\star \ \ 8x^2-6x+1=0\ \ ,\ \ D/4=(b/2)^2-ac=3^2-8=1\ ,\\\\x_1=\dfrac{3-1}{8}=\dfrac{1}{4}\ \ ,\ \ \ x_2= \dfrac{3+1}{8}=\dfrac{1}{2}\ \ \star[/tex]
[tex]b)\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x(-x-\dfrac{3}{4})=\dfrac{1}{8}\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}-x^2-\dfrac{3}{4}\, x-\dfrac{1}{8}=0\\y=-x-\dfrac{3}{4}\end{array}\right[/tex]
[tex]\left\{\begin{array}{l}8x^2+6x+1=0\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x_1=-\dfrac{1}{4}\ ,\ x_2=-\dfrac{1}{2}\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}\ x_1=-\dfrac{1}{4}\ ,\ x_2=-\dfrac{1}{2}\\y_1=-\dfrac{1}{2}\ ,\ y_2=-\dfrac{1}{4}\end{array}\right\\\\\\\star \ \ 8x^2+6x+1=0\ \ ,\ \ D/4=(b/2)^2-ac=3^2-8=1\ ,\\\\x_1=\dfrac{-3-1}{8}=-\dfrac{1}{2}\ \ ,\ \ \ x_2= \dfrac{-3+1}{8}=-\dfrac{1}{4}\ \ \star[/tex]
[tex]\bf Otvet:\ \Big(\ \dfrac{1}{4}\ ;\ \dfrac{1}{2}\ \Big)\ ,\ \Big(\ \dfrac{1}{2}\ ;\ \dfrac{1}{4}\ \Big)\ ,\ \Big(-\dfrac{1}{4}\ ;-\dfrac{1}{2}\ \Big)\ ,\ \Big(-\dfrac{1}{2}\ ;-\dfrac{1}{4}\ \Big)\ .[/tex]
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Answers & Comments
Ответ:
[tex]\left\{\begin{array}{l}xy=\dfrac{1}{8}\\2x^2+2y^2=\dfrac{5}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\2(x^2+y^2)=\dfrac{5}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\ \ \ \Big|\cdot 2\\x^2+y^2=\dfrac{5}{16}\end{array}\right\ \ \oplus[/tex]
Умножим на 2 первое уравнение и прибавим ко второму уравнению
и применим формулу квадрата суммы: [tex](x+y)^2=x^2+y^2+2xy[/tex] .
[tex]\left\{\begin{array}{l}xy=\dfrac{1}{8}\\x^2+y^2+2xy=\dfrac{5}{16}+\dfrac{2}{8}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\(x+y)^2=\dfrac{5}{16}+\dfrac{1}{4}\end{array}\right\\\\\\\left\{\begin{array}{l}xy=\dfrac{1}{8}\\(x+y)^2=\dfrac{9}{16}\end{array}\right\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=\pm \dfrac{3}{4}\end{array}\right[/tex]
[tex]a)\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x(-x+\dfrac{3}{4})=\dfrac{1}{8}\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}-x^2+\dfrac{3}{4}\, x-\dfrac{1}{8}=0\\y=-x+\dfrac{3}{4}\end{array}\right[/tex]
[tex]\left\{\begin{array}{l}8x^2-6x+1=0\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x_1=\dfrac{1}{4}\ ,\ x_2=\dfrac{1}{2}\\y=-x+\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}\ x_1=\dfrac{1}{4}\ ,\ x_2=\dfrac{1}{2}\\y_1=\dfrac{1}{2}\ ,\ y_2=\dfrac{1}{4}\end{array}\right\\\\\\\star \ \ 8x^2-6x+1=0\ \ ,\ \ D/4=(b/2)^2-ac=3^2-8=1\ ,\\\\x_1=\dfrac{3-1}{8}=\dfrac{1}{4}\ \ ,\ \ \ x_2= \dfrac{3+1}{8}=\dfrac{1}{2}\ \ \star[/tex]
[tex]b)\ \ \left\{\begin{array}{l}xy=\dfrac{1}{8}\\x+y=-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x(-x-\dfrac{3}{4})=\dfrac{1}{8}\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}-x^2-\dfrac{3}{4}\, x-\dfrac{1}{8}=0\\y=-x-\dfrac{3}{4}\end{array}\right[/tex]
[tex]\left\{\begin{array}{l}8x^2+6x+1=0\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}x_1=-\dfrac{1}{4}\ ,\ x_2=-\dfrac{1}{2}\\y=-x-\dfrac{3}{4}\end{array}\right\ \ \left\{\begin{array}{l}\ x_1=-\dfrac{1}{4}\ ,\ x_2=-\dfrac{1}{2}\\y_1=-\dfrac{1}{2}\ ,\ y_2=-\dfrac{1}{4}\end{array}\right\\\\\\\star \ \ 8x^2+6x+1=0\ \ ,\ \ D/4=(b/2)^2-ac=3^2-8=1\ ,\\\\x_1=\dfrac{-3-1}{8}=-\dfrac{1}{2}\ \ ,\ \ \ x_2= \dfrac{-3+1}{8}=-\dfrac{1}{4}\ \ \star[/tex]
[tex]\bf Otvet:\ \Big(\ \dfrac{1}{4}\ ;\ \dfrac{1}{2}\ \Big)\ ,\ \Big(\ \dfrac{1}{2}\ ;\ \dfrac{1}{4}\ \Big)\ ,\ \Big(-\dfrac{1}{4}\ ;-\dfrac{1}{2}\ \Big)\ ,\ \Big(-\dfrac{1}{2}\ ;-\dfrac{1}{4}\ \Big)\ .[/tex]