Ta có :
\(a^2+b^2=2ab\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
\(a=b\)
Vậy ĐPCM
\(a^2+b^2-2ab=0\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\left(dpcm\right)\)
\(a^2+b^2=2ab\)
\(\Rightarrow a^2-2ab+b^2=0\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a-b=0\)
\(\Rightarrow a=b\)
Chúc bạn học tốt.
Ta có: \(a^2+b^2=2ab\)
\(\Rightarrow a^2-2ab+b^2=0\)
\(\left(a-b\right)^2=0\)
\(a-b=0\)
\(\Rightarrow a=b\)
=.= hok tốt!!
\(a^2+b^2=2ab\)
\(\Rightarrow a^2-2ab+b^2=0\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a=b\)
a2+b2=2ab
=> a2-2ab+b2=0
=> (a-b)2=0 (HDT số 2)
=> a-b=0
=>a=b
