Question

biết x , y thỏa mãn /x+1/ + / x-y+2 / = 0 . khi đó \(x^2+y^2+1=?\)

( / la gia tri tuyet doi )

Answer 1

Vì \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left|x-y+2\right|\ge0\end{cases}\Rightarrow}\left|x+1\right|+\left|x-y+2\right|\ge0\)

Theo đề bài: \(\left|x+1\right|+\left|x-y+2\right|=0\Leftrightarrow\hept{\begin{cases}x+1=0\\x-y+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

=> \(x^2+y^2+1=1+1+1=3\)

Answer 2

\(!x+1!+!x-y+2!=0\\ \) khi

\(\hept{\begin{cases}x+1=0\\x-y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\Rightarrow x^2+y^2+1=3}\)

Answer 3

Vì \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left|x-y+2\right|\ge0\end{cases}\Rightarrow}\left|x+1\right|+\left|x-y+2\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+1=0\\x-y+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

=> \(x^2+y^2+1=\left(-1\right)^2+1^2+1=1+1+1=3\)

Vậy .......................