Toán Lớp 7: x2 + 5y2 + 2x – 4xy – 10y + 14 > 0 với mọi x, y.
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Toán Lớp 7: x2 + 5y2 + 2x – 4xy – 10y + 14 > 0 với mọi x, y.
Comments ( 1 )
Ta có: A=x^2+5y^2+2x-4xy-10y+14
=x^2-4xy+4y^2+(2x-4y)+(y^2-6y+9)+5
=(x-2y)^2+2(x-2y)+1(y-3)^2+4
=(x-2y+1)^2+(y-3)^2+4
Vì: (x-2y+1)^2≥0
(y-3)^2≥0
→(x-2y+1)^2+(y-3)^2≥0
→(x-2y+1)^2+(y-3)^2+4≥4
→A≥4
Dấu “=” xảy ra khi:
+) y-3=0
→y=3
+) Thay y=3 vào x-2y+1=0, ta có:
x-2.3+1=0
→x-6+1=0
→x-5=0
→x=5
Vậy (x;y)=(5;3)