Tìm x;y;z biết : x+y/5 = x-y/1 = xy/2

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Tìm x;y;z biết :
x+y/5 = x-y/1 = xy/2

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Eirian 4 years 2020-11-02T03:14:19+00:00 1 Answers 62 views 0

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  1. Helga
    0
    2020-11-02T03:15:34+00:00
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    Đáp án: $(x,y)\in\{(0,0), (1,\dfrac23)\}$

    Giải thích các bước giải:

    Ta có:
    $\dfrac{x+y}{5}=\dfrac{x-y}{1}=\dfrac{xy}{2}=\dfrac{x+y+x-y}{5+1}=\dfrac{2x}{6}=\dfrac{x}{3}$
    $\to \dfrac{xy}{2}=\dfrac{x}{3}$
    $\to 3xy=2x$
    $\to 3xy-2x=0$
    $\to x(3y-2)=0$
    $\to x=0$ hoặc $3y-2=0\to 3y=2\to y=\dfrac23$
    Nếu $x=0$
    $\to \dfrac{0+y}{5}=\dfrac{0-y}{1}\to \dfrac{y}{5}=\dfrac{-y}{1}$
    $\to -y=5y$
    $\to 6y=0$
    $\to y=0$
    $\Rightarrow (x,y)=(0,0)$
    Nếu $y=\dfrac23$
    $\to \dfrac{x+\dfrac23}{5}=\dfrac{x-\dfrac23}{1}=\dfrac{x\cdot\dfrac23}{2}$
    $\to \dfrac{x+\dfrac23}{5}=\dfrac{x-\dfrac23}{1}$
    $\to x+\dfrac23=5(x-\dfrac23)$
    $\to x+\dfrac23=5x-\dfrac{10}{3}$
    $\to 5x-x=\dfrac{10}{3}+\dfrac{2}{3}$
    $\to 4x=\dfrac{12}{3}=4$
    $\to x=1$

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )