a) Vì \(\hept{\begin{cases}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{cases}}\)mà \(\left|x\right|+\left|x+2\right|=0\)nên \(\hept{\begin{cases}\left|x\right|=0\\\left|x+2\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}\)(vô lý)
[x]+[x+2]=0
=>[x]=0 =>x=0
[x+2]=0
b)
\(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\)
Điều kiện \(x\ge0\)nên
\(x\left(x^2-\frac{5}{4}\right)=x\)
\(\Leftrightarrow x^2-\frac{5}{4}=x:x=1\)
\(\Leftrightarrow x^2=\frac{9}{4}\)
\(\Rightarrow x=\sqrt{\frac{9}{4}}=\frac{3}{2}\)