\(\left|x\right|+\left|x+2\right|=\left|-x\right|+\left|x+2\right|mà:\left\{{}\begin{matrix}\left|-x\right|\ge-x\\\left|x+2\right|\ge x+2\end{matrix}\right.\Rightarrow\left|x\right|+\left|x+2\right|\ge-x+x+2=2>0\Rightarrow x\in\varnothing\)
Mị thấy sai sai nên mị có ý kiến khác.
a) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left|x\right|\ge0\) với mọi x
\(\left|x+2\right|\ge0\) với mọi x
\(\Rightarrow\left|x\right|+\left|x+2\right|\ge0\) với mọi x
Mà \(\left|x\right|+\left|x+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=0\\\left|x+2\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{0;-2\right\}\)
b) \(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\left(ĐK:x\ge0\right)\)
\(\Rightarrow\left[{}\begin{matrix}x\left(x^2-\frac{5}{4}\right)=x\\x\left(x^2-\frac{5}{4}\right)=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\left(x^2-\frac{5}{4}\right)-x=0\\x\left(x^2-\frac{5}{4}\right)+x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x\left(x^2-\frac{5}{4}-1\right)=0\\x\left(x^2-\frac{5}{4}+1\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\frac{9}{4}\\x^2=\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\frac{3}{2}\left(\text{vì }x\ge0\right)\\x=\frac{1}{2}\left(\text{vì }x\ge0\right)\end{matrix}\right.\)
Vậy \(x\in\left\{0;\frac{3}{2};\frac{1}{2}\right\}\)
