1 câu trả lời
1a-1b=1a-b
⇔b-aab=-1b-a
⇒(b-a)²=-ab
⇔a²-2ab+b²=-ab
⇔a²-ab+b²=0
⇔a²-2.a.b2+b²4+3b²4=0
⇔(a-b2)²+3b²4=0
Ta có
(a-b2)²+3b²4≥0
Dấu = xảy ra khi : a-b2=0,3b²4=0
⇒a=b=0
1a-1b=1a-b
⇔b-aab=-1b-a
⇒(b-a)²=-ab
⇔a²-2ab+b²=-ab
⇔a²-ab+b²=0
⇔a²-2.a.b2+b²4+3b²4=0
⇔(a-b2)²+3b²4=0
Ta có
(a-b2)²+3b²4≥0
Dấu = xảy ra khi : a-b2=0,3b²4=0
⇒a=b=0