Đáp án: $ (x,y)\in\{(1,-5),(1,5),(-1,-5),(-1,5),(5,1),(5,-1),(-5,1),(-5,1)\}$
Giải thích các bước giải:
Ta có :
$|x|+|y|=6\to |y|=6-|x|$
$\to |y|^2=(6-|x|)^2$
$\to y^2=36-12|x|+|x|^2$
$\to 26-x^2=36-12|x|+x^2, x^2+y^2=26$
$\to 2x^2-12|x|+10=0$
$\to x^2-6|x|+5=0$
$\to |x|^2-6|x|+5=0$
$\to (|x|-1)(|x|-5)=0$
$\to |x|\in\{1,5\}$
$+)|x|=1\to x\in\{1,-1\}$
$\to |y|=6-|x|=5\to y\in\{-5,5\}$
$\to (x,y)\in\{(1,-5),(1,5),(-1,-5),(-1,5)\}$
$+)|x|=5\to x\in\{5,-5\}$
$\to |y|=6-|x|=1\to y\in\{1,-1\}$
$\to (x,y)\in\{(5,1),(5,-1),(-5,1),(-5,1)\}$
