Giải thích các bước giải:
Ta có : $3(a^2+b^2)-7(a+b)=-4$
$\to 3(a+b)^2-6ab-7(a+b)+4=0$
$\to 3(a+b)^2-7(a+b)+4=6ab\le 6.\dfrac{(a+b)^2}{4}=\dfrac{3(a+b)^2}{2}$
$\to \dfrac{3(a+b)^2}{2}-7(a+b)+4\le 0$
$\to 3(a+b)^2-14(a+b)+8\le 0$
$\to (3(a+b)-2)(a+b-4)\le 0$
$\to \dfrac 23\le a+b\le 4$
$\to a+b\in\{0,1,2,3\}$
$\to ab\in\{\dfrac 23, 0,\dfrac 13, \dfrac 53\}$ vì $3(a+b)^2-7(a+b)+4=6ab$
$\to a+b=1, ab=0\to (a,b)\in\{(0,1), (1,0)\}$
