b: \(\left|2-x\right|+\left|x^2-4\right|=0\)
=>\(\left|x-2\right|+\left|x-2\right|\cdot\left|x+2\right|=0\)
=>\(\left|x-2\right|\left(\left|x+2\right|+1\right)=0\)
mà \(\left|x+2\right|+1>=1>0\forall x\)
nên |x-2|=0
=>x-2=0
=>x=2
a: \(\left|x-8\right|+\left|x-5\right|=3x\)(1)
TH1: x<5
(1) sẽ trở thành 5-x+8-x=3x
=>-2x+13=3x
=>-5x=-13
=>x=2,6(nhận)
TH2: 5<=x<8
(1) sẽ trở thành x-5+8-x=3x
=>3x=3
=>x=1(loại)
TH3: x>=8
(1) sẽ trở thành x-5+x-8=3x
=>3x=2x-13
=>x=-13(loại)
Vậy: x=2,6
c: \(\left(x+1\right)^2-\left|x+1\right|=0\)
=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|=0\)
=>\(\left|x+1\right|\left(\left|x+1\right|-1\right)=0\)
=>\(\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)
d: \(\left|x^2+2|x-\dfrac{1}{2}|\right|=x^2+2\)
=>\(x^2+2\left|x-\dfrac{1}{2}\right|=x^2+2\)
=>\(\left|x-\dfrac{1}{2}\right|=1\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)