$x^{2} – 5xy – x + 5y + 1 = 0$

1 bình luận về “$x^{2} – 5xy – x + 5y + 1 = 0$”

1. x^2-5xy-x+5y+1=0

   =>x(x-5y)-(x-5y)=-1

   =>(x-1)(x-5y)=-1

   => \(\left[ \begin{array}{l}\begin{cases}x-1=1\\x-5y=-1\\\end{cases}\\\begin{cases}x-1=-1\\x-5y=1\\\end{cases}\end{array} \right.\) => \(\left[ \begin{array}{l}\begin{cases}x=2\\y=\dfrac{3}{5}\\\end{cases}\\\begin{cases}x=0\\x=-\dfrac{1}{5}\\\end{cases}\end{array} \right.\) 

     =>(x;y)={(2;3/5);(0;-1/5)}

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