Question

\(\text{Cho }a+b+c=0\text{ thỏa mãn }a^2=\left(2a+2b+2\right)\left(a+c-1\right)\)

Answer 1

\(a=-b-c\)

\(\Rightarrow\left(-b-c\right)^2=\left(2a+2b+2\right)\left(a+c-1\right)\)

\(\Rightarrow b^2+2bc+c^2=2a^2+2ac-2a+2ab+2bc-2b+2a+2c-2\)

\(\Leftrightarrow b^2+2b+1+c^2-2c+1=2a^2+2ab+2ac\)

\(\Leftrightarrow\left(b+1\right)^2+\left(c-1\right)^2=2a\left(a+b+c\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}b+1=0\\c-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0\\b=-1\\c=1\end{matrix}\right.\) \(\Rightarrow P=-1\)

Answer 2

\(\text{Tính }P=ab+bc+ca\)