a) ĐK: x \(\ge\) 1
Khi đó: |(x-1)(x2-3)| = x-1
=> \(\left\{{}\begin{matrix}\left(x-1\right)\left(x^2-3\right)=x-1\\\left(x-1\right)\left(x^2-3\right)=1-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(x-1\right)\left(x^2-3-1\right)=0\\\left(x-1\right)\left(x^2-3+1\right)=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-1=0\\x^2-4=0\\x^2-2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\left(t.m\right)\\x=2\left(t.m\right)\\x=-2\left(k.t.m\right)\\x=\sqrt{2}\left(t.m\right)\end{matrix}\right.\)
Vậy x= 1 hoặc x = 2 hoặc x = \(\sqrt{2}\)
b) |x+2| + 2x = 1
=> |x+2| = 1 - 2x
=> \(\left\{{}\begin{matrix}x+2=1-2x\\x+2=-1+2x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x=-1\\-x=-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\left(t.m\right)\\x=3\left(k.t.m\right)\end{matrix}\right.\)
Vậy x = \(-\dfrac{1}{3}\)
