Giải hệ phương trình : $left { {{frac{x^2}{(y+1)^2}+frac{y^2}{(x+1)^2}=frac{2}{1}\3xy=x+y+1 } atop {}} ight.$

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Giải hệ phương trình : $left { {{frac{x^2}{(y+1)^2}+frac{y^2}{(x+1)^2}=frac{2}{1}\3xy=x+y+1 } atop {}}
ight.$

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Aaliyah 1 giờ 2021-09-12T00:24:04+00:00 1 Answers 0 views 0

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    2021-09-12T00:25:59+00:00

    \(\begin{array}{l}
    \left\{ \begin{array}{l}
    \dfrac{{{x^2}}}{{{{(y + 1)}^2}}} + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} = 2(1)\\
    3xy = x + y + 1(2)
    \end{array} \right.\\
    (1) \Leftrightarrow \dfrac{{{x^2}}}{{{{(y + 1)}^2}}} – 1 + \dfrac{{{y^2}}}{{{{(x + 1)}^2}}} – 1 = 0\\
    \Leftrightarrow \left( {\dfrac{x}{{y + 1}} – 1} \right)\left( {\dfrac{x}{{y + 1}} + 1} \right) + \left( {\dfrac{y}{{x + 1}} – 1} \right)\left( {\dfrac{y}{{x + 1}} + 1} \right) = 0\\
    \Leftrightarrow \dfrac{{x – y – 1}}{{y + 1}}\frac{{x + y + 1}}{{y + 1}} + \dfrac{{y – x – 1}}{{x + 1}}\dfrac{{y + x + 1}}{{x + 1}} = 0\\
    \Leftrightarrow \left( {x + y + 1} \right)\left[ {\dfrac{{x – y – 1}}{{{{(y + 1)}^2}}} + \dfrac{{y – x – 1}}{{{{(x + 1)}^2}}}} \right] = 0\\
    \Rightarrow \left[ \begin{array}{l}
    x + y + 1 = 0(*)\\
    \dfrac{{x – y – 1}}{{{{(y + 1)}^2}}} + \dfrac{{y – x – 1}}{{{{(x + 1)}^2}}} = 0
    \end{array} \right.(**)
    \end{array}\)
    Thay (*) vào (2) ta có:
    \(\begin{array}{l}
    (2) \Rightarrow 3xy = 0\\
    \Rightarrow \left[ \begin{array}{l}
    x = 0\\
    y = 0
    \end{array} \right.\\
    x = 1 \Rightarrow (1) \Rightarrow \dfrac{1}{{{{(y + 1)}^2}}} + \dfrac{{{y^2}}}{{{{(1 + 1)}^2}}} = 2\\
    \Rightarrow 4 + {y^2}{(y + 1)^2} = 8{(y + 1)^2}
    \end{array}\)
    \( \Rightarrow (y + 2)({y^3} – 7x – 2) = 0\)
    Suy ra có 1 nghiệm y=2

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