Tìm x,y : a,(x-1)²+(y-2)²=0 b,(x+2)²+(y-5)²=0 c,(x-2)²+(x+y-2)²=0 d, (x+1)²+(x-y-3)²=0 Giúp mình với ! Cảm ơn các bạn nhiều !

$a),(x-1)^2+(y-2)^2=0$

 Vì : $(x-1)^2 ≥ 0$ $∀$ $x$

       $(y-2)^2  ≥ 0$ $∀$ $y$

$⇒ (x-1)^2 + (y-2)^2 ≥ 0$

$⇒$ \(\left[ \begin{array}{l}x-1=0\\y-2=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=1\\y=2\end{array} \right.\) 

   Vậy `(x;y)=(1;2)`

$b) (x+2)^2+(y-5)^2=0$

 Vì : $(x+2)^2 ≥ 0$ $∀$ $x$

       $(y-5)^2  ≥ 0$ $∀$ $y$

$⇒ (x+2)^2+(y-5)^2 ≥ 0$

$⇒$ \(\left[ \begin{array}{l}x+2=0\\y-5=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=-2\\y=5\end{array} \right.\) 

   Vậy `(x;y)=(-2;5)`

$c) (x-2)^2+(x+y-2)^2=0$

 Vì : $(x-2)^2 ≥ 0$ $∀$ $x$

       $(x+y-2)^2  ≥ 0$ $∀$ $y$

$⇒ (x-2)^2+(x+y-2)^2 ≥ 0$

$⇒$ \(\left[ \begin{array}{l}x-2=0\\x+y=2\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=2\\y=0\end{array} \right.\) 

   Vậy `(x;y)=(2;0)`

$d) (x+1)^2+(x-y-3)^2=0$

 Vì : $(x+1)^2 ≥ 0$ $∀$ $x$

       $(x-y-3)^2  ≥ 0$ $∀$ $y$

$⇒ (x+1)^2+(x-y-3)^2 ≥ 0$

$⇒$ \(\left[ \begin{array}{l}x+1=0\\x-y=3\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=-1\\y=-4\end{array} \right.\) 

   Vậy `(x;y)=(-1;-4)`