Question

(x+3)² +(x-1)² = 0

Answer 1

Vì (x+3)² \(\ge0,\forall x\) và (x-1)² \(\ge0,\forall x\)

Mà theo bài ra : \(\left(x+3\right)^2+\left(x-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\) => phương trình vô nghiệm

Answer 2

Với mọi giá trị của x ta có:

\(\left(x+3\right)^2+\left(x-1\right)^2\ge0\)

Để \(\left(x+3\right)^2+\left(x-1\right)^2=0\) thì

\(\left\{{}\begin{matrix}\left(x+3\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\Rightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

Chúc bạn học tốt!!!

Answer 3

\(\left(x+3\right)^2+\left(x-1\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x+3\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\)=> \(\left(x+3\right)^2+\left(x-1\right)^2\ge0\)

Khi \(\left\{{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-3\\x=0+1\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)