Question

Cho \(a,b,c\in R\) sao cho \(a+b+c=0\) và \(a^2=2\left(a+c+1\right)\left(a+b-1\right)\)

Tính \(A=a^2+b^2+c^2\)

Answer 1

đề có chút kì lạ ấy

...

hóng sol đẹp

Answer 2

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

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

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

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

\(\Leftrightarrow b^2+2bc+c^2=2bc+2b-2c-2\)

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

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

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

\(\Rightarrow A=2\)