\(A=\left(x+1\right)^2+\left(y-2\right)^2+9\)

Có: \(\left(x+1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\left(x+1\right)^2+\left(y-2\right)^2+9\ge9\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\) .

Vậy: \(Min_A=9\) tại \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

cau nay cx hoi dc

\(\left|x-2\right|=2-x\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\2-x=0\end{cases}}\Rightarrow x=\hept{\begin{cases}2\\2\end{cases}}\)

Vậy x = 2 

tu bai tho to laong e hieu the nao la tri lam trai