Em tham khảo:
Ta có ĐKXĐ $x+y$ khác 0
$x^{2}-2y^2=xy$
⇔$x^{2}-y^2-y^2-xy=0$
⇔$(x-y)(x+y)-y(x+y)=0$
⇔$(x+y)(x-2y)=0$
⇔\(\left[ \begin{array}{l}x=2y\\x+2y(loại)\end{array} \right.\)
Với $2y=x$
⇒$P=\dfrac{2y-y}{2y+y}=$ $\dfrac{y}{3y}=$ $\dfrac{1}{3}$
Học tốt
Đáp án:
$P =\dfrac13$
Giải thích các bước giải:
$\quad x^2 - 2y^2 = xy$
$\to x^2 - y^2 - y^2 - xy = 0$
$\to (x-y)(x+y) - y(x+y)=0$
$\to (x+y)(x-2y)=0$
$\to \left[\begin{array}{l}x + y = 0\quad (loại)\\x - 2y = 0\quad (nhận)\end{array}\right.$
$\to x = 2y$
Ta được:
$\quad P = \dfrac{x-y}{x+y}$
$\to P =\dfrac{2y - y}{2y + y}$
$\to P =\dfrac{y}{3y}$
$\to P =\dfrac13$
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