Question

Tìm x,y biết x²+y²-2x+4y+5=0

Answer 1

Ta có: \(x^2+y^2-2x+4y+5=0\)

<=> \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

<=> \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)

=> \(\left[\begin{array}{nghiempt}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x-1=0\\y+2=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=1\\y=-2\end{array}\right.\)

Vậy x=1 ; y=-2