Question

[Nguyễn Công Minh Hoàng]

Tìm a; b; c biết:

a)\(a^2+2b^2-2ab+2a-4b+2=0\)

b)\(a^2+5b^2-4ab+2a-6b+2=0\)

Answer 1

[Trần Phúc Khang]

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

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

Mà \(\left(a-b+1\right)^2\ge0,\left(b-1\right)^2\ge0\)

=> \(\hept{\begin{cases}a-b+1=0\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}}\)

b,Tương tự 

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

=>\(\hept{\begin{cases}a=1\\b=1\end{cases}}\)