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\(\Leftrightarrow a^2+b^2-2ab+2a-2b=63\)
\(\Leftrightarrow\left(b-a\right)^2-2\left(b-a\right)-63=0 \)
\(\Leftrightarrow\left(b-a\right)^2-9\left(b-a\right)+7\left(b-a\right)-63=0\)
\(\Leftrightarrow\left(b-a\right)\left(b-a-9\right)+7\left(b-a-9\right)=0\)
\(\Leftrightarrow\left(b-a-9\right)\left(b-a+7\right)=0\)
\(\Leftrightarrow b-a-9=0\) hoặc \(b-a+7=0\)
\(\Leftrightarrow b-a=9\) hoặc \(b-a=-7\left(l\right)\) vì b > a
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![](https://rs.olm.vn/images/avt/0.png?1311)
Thay \(c=2-\left(a+b\right)\Leftrightarrow P=2ab+c\left(a+b\right)=2ab+\left(a+b\right)\left[2-\left(a+b\right)\right]\)
\(=2ab+2\left(a+b\right)-a^2-b^2-2ab=2\left(a+b\right)-a^2-b^2=2-\left(a-1\right)^2-\left(b-1\right)^2\)
Mà \(\hept{\begin{cases}\left(a-1\right)^2\\\left(b-1\right)^2\end{cases}\ge0\forall a,b\inℝ\Rightarrow P=2-\left(a-1\right)^2-\left(b-1\right)^2\le2}\)
Dấu ''='' xảy ra \(\Leftrightarrow\) \(a=b=1\rightarrow c=0\)