Question

cho a^2+b^2+1=a+b+a*b. Chung minh a=b=1

 

Answer 1

\(a^2+b^2+1=a+b+ab\Leftrightarrow2a^2+2b^2+2=2a+2b+2ab\)

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

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

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\a-1=0\\b-1=0\end{cases}\Leftrightarrow}a=b=1\)