A=-5 thì x-|x-2|=-5
=>|x-2|=x+5
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(x-2-x-5\right)\left(x-2+x+5\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{3}{2}\)
A=2 thì x-|x-2|=2
=>|x-2|=x-2
=>x-2>=0
hay x>=2
\(A=-5=>x-\left|x-2\right|=-5\) \((*)\)
\(+) TH1:x >=2\)
\(=>x-2>=0=>\left|x-2\right|=x-2\)
\((*)\)\(=>x-\left(x-2\right)=-5< =>2=-5\) \(( Vô\ lí )\)
\(=>x\in\varnothing\)
\(+)TH2:x<2\)
\(=>x-2< 0=>\left|x-2\right|=-\left(x-2\right)\)
\((*)\)\(=>x-\left[-\left(x-2\right)\right]=-5< =>x+x-2=-5< =>2x=-3< =>x=-\dfrac{3}{2}\left(TMDK\right)\)
\(Vậy\ để\ A=-5\ thì\ x=-3/2\)
\(A=2=>x-\left|x-2\right|=2\)\((**)\)
\(+)TH1:x>=2\)
\((**)\)\(=>x-\left(x-2\right)=2< =>2=2\)\((Luôn\ đúng)\)
\(=>x>=2\)
\(+)TH2:x<2\)
\(=>x-\left[-\left(x-2\right)\right]=2< =>2x-2=2< =>x=2\left(KTMDK\right)\)
\(=>x\in\varnothing\)
\(Vậy\ để\ A=2\ thì \ x>=2\)
