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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Tìm x,y bt a) 9x^2+12x+4y^2+8y+8=0 b) 4x^2+9y^2-10x-12y+29=0 c) x^2+4y^2+4x-4y+5=0 d) 4x^2-12x+y^2-4y+13=0 MỌI NGƯỜI ƠI GIÚP MIK VS

Toán Lớp 9: Tìm x,y bt
a) 9x^2+12x+4y^2+8y+8=0
b) 4x^2+9y^2-10x-12y+29=0
c) x^2+4y^2+4x-4y+5=0
d) 4x^2-12x+y^2-4y+13=0
MỌI NGƯỜI ƠI GIÚP MIK VS MIK ĐG CẦN GẤP Ạ

Comments ( 1 )

  1. Lời giải và giải thích chi tiết:
    Ta có:
    \(\begin{array}{l}
    a,\\
    9{x^2} + 12x + 4{y^2} + 8y + 8 = 0\\
     \Leftrightarrow \left( {9{x^2} + 12x + 4} \right) + \left( {4{y^2} + 8y + 4} \right) = 0\\
     \Leftrightarrow \left[ {{{\left( {3x} \right)}^2} + 2.3x.2 + {2^2}} \right] + \left[ {{{\left( {2y} \right)}^2} + 2.2y.2 + {2^2}} \right] = 0\\
     \Leftrightarrow {\left( {3x + 2} \right)^2} + {\left( {2y + 2} \right)^2} = 0\\
    {\left( {3x + 2} \right)^2} \ge 0,\,\,\,\forall x\\
    {\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall y\\
     \Rightarrow {\left( {3x + 2} \right)^2} + {\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
     \Rightarrow \left\{ \begin{array}{l}
    {\left( {3x + 2} \right)^2} = 0\\
    {\left( {2y + 2} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    3x + 2 = 0\\
    2y + 2 = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x =  – \dfrac{2}{3}\\
    y =  – 1
    \end{array} \right.\\
    c,\\
    {x^2} + 4{y^2} + 4x – 4y + 5 = 0\\
     \Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left( {4{y^2} – 4y + 1} \right) = 0\\
     \Leftrightarrow \left( {{x^2} + 2.x.2 + {2^2}} \right) + \left[ {{{\left( {2y} \right)}^2} – 2.2y.1 + {1^2}} \right] = 0\\
     \Leftrightarrow {\left( {x + 2} \right)^2} + {\left( {2y – 1} \right)^2} = 0\\
    {\left( {x + 2} \right)^2} \ge 0,\,\,\,\forall x\\
    {\left( {2y – 1} \right)^2} \ge 0,\,\,\,\forall y\\
     \Rightarrow {\left( {x + 2} \right)^2} + {\left( {2y – 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
     \Rightarrow \left\{ \begin{array}{l}
    {\left( {x + 2} \right)^2} = 0\\
    {\left( {2y – 1} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x + 2 = 0\\
    2y – 1 = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x =  – 2\\
    y = \dfrac{1}{2}
    \end{array} \right.\\
    d,\\
    4{x^2} – 12x + {y^2} – 4y + 13 = 0\\
     \Leftrightarrow \left( {4{x^2} – 12x + 9} \right) + \left( {{y^2} – 4y + 4} \right) = 0\\
     \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} – 2.2x.3 + {3^2}} \right] + \left[ {{y^2} – 2.y.2 + {2^2}} \right] = 0\\
     \Leftrightarrow {\left( {2x – 3} \right)^2} + {\left( {y – 2} \right)^2} = 0\\
    {\left( {2x – 3} \right)^2} \ge 0,\,\,\,\forall x\\
    {\left( {y – 2} \right)^2} \ge 0,\,\,\,\forall y\\
     \Rightarrow {\left( {2x – 3} \right)^2} + {\left( {y – 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
     \Rightarrow \left\{ \begin{array}{l}
    {\left( {2x – 3} \right)^2} = 0\\
    {\left( {y – 2} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    2x – 3 = 0\\
    y – 2 = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x = \dfrac{3}{2}\\
    y = 2
    \end{array} \right.
    \end{array}\)
    Em xem lại đề câu b nhé!

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222-9+11+12:2*14+14 = ? ( )