Tìm x, y biết: 10×2- 2xy+ y2+ 24x+ 16=0

2 bình luận về “Tìm x, y biết: 10×2- 2xy+ y2+ 24x+ 16=0”

1. 10x^2 – 2xy + y^2 + 24x + 16 = 0

   => (x^2 – 2xy + y^2) + (9x^2 + 24x + 16) = 0

   => (x – y)^2 + (3x + 4)^2 = 0

   Vì {((x -y)^2 \ge 0),((3x + 4)^2 ‘ge 0):}

   => (x – y)^2 + (3x + 4)^2 \ge 0

   Dấu “=” xảy ra <=> {((x -y)^2 = 0),((3x + 4)^2 = 0):}

   <=> {(x – y = 0),(3x + 4 = 0):}

   <=> {(x = y),(x = -4/3):}

   => x = y = -4/3

   Vậy x = y = -4/3

   $#duong612009$

   Trả lời
2. 10x^2 – 2xy + y^2 + 24x + 16 = 0

   <=> x^2 + 9x^2 – 2xy + y^2 + 24x + 16 = 0

   <=> (x^2 – 2xy + y^2) + (9x^2 + 24x + 16) = 0

   <=> (x – y)^2 + (3x + 4)^2 = 0

   Mà (x – y)^2 ; (3x + 4)^2 ≥ 0 ∀ x,y

   =>$\begin{cases} (x – y)^2 =0\\ (3x + 4)^2=0 \end{cases}$

   <=>$\begin{cases} x – y =0\\ 3x + 4=0 \end{cases}$

   <=> $\begin{cases} y =\dfrac{-4}{3}\\ x=\dfrac{-4}{3} \end{cases}$

   Vậy x = -4/3 ; y = -4/3

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