

a)Ta có:
x2+xy+y2=2x+y⇔x2−2x+xy+y2−y=0⇔2x2−4x+2xy+2y2−2y=0⇔(x2−4x+4)+(x2+2xy+y2)+(y2−2y+1)=5⇔(x−2)2+(x+y)2+(y−1)2=5⇒(x−2)2≤5⇔∣x−2∣=0hoặc∣x−2∣=1
+ |x-2|=0
⇒x=2⇒(y+2)2+(y−1)2=5⇒2y2+2y+5=5⇒2y2+2y=0⇒2y(y+1)=0⇒y=0hoặcy=1
+ |x-2|=1
⇒x=1hoặcx=3
* x=1
⇒(y+1)2+(y−1)2=1⇒y2+2=1⇒y2=−1(loại)
* x=3
⇒(y+3)2+(y−1)2=1⇔2y2+4y+10=1⇒2(y+1)2=−7(loại)
Vậy
(x,y)=(2,−1);(2,0)
b)
x2−2xy+5y2=y+1⇔(x−y)2+4y2−y=1⇔16(x−y)2+64y2−16y=16⇔16(x−y)2+(8y−1)2=17
⇒16(x−y)2≤17⇒∣x−y∣=0hoặc∣x−y∣=1
+ |x-y|=0
⇒{x=y(8y−1)2=17(loại)
+ |x-y|=1
⇒(8y−1)2=1⇔64x2−16x=0⇔16x(4x−1)=0⇒x=0(t/m)hoặcx=41(loại)
Thay vào ta có:
∣x∣=1⇒x=±1
Vậy
(x,y)=(1,0);(−1,0)