a: \(A=3-\left|3-x\right|\)
=3-|x-3|
TH1: x>=3
=>x-3>=0
A=3-|x-3|
=3-(x-3)
=3-x+3
=6-x
TH2: x<3
=>x-3<0
A=3-|x-3|
=3-(3-x)
=3-3+x
=x
b: \(B=\left|x-6\right|+\left|6-x\right|-2\)
=2|x-6|-2
TH1: x>=6
=>x-6>=0
B=2|x-6|-2
=2(x-6)-2
=2x-12-2
=2x-14
TH2: x<6
=>x-6<0
B=2|x-6|-2
=2(6-x)-2
=12-2x-2
=10-2x
c: \(C=\left|-x-1\right|+\left|-x-5\right|-x\)
\(=\left|x+5\right|+\left|x+1\right|-x\)
TH1: x<-5
=>x+5<0; x+1<0
\(C=\left|x+5\right|+\left|x+1\right|-x\)
=-x-5-x-1-x
=-3x-6
TH2: -5<=x<-1
=>x+5>=0; x+1<0
\(C=\left|x+5\right|+\left|x+1\right|-x\)
=x+5-x-1-x
=-x+4
TH3: x>=-1
=>x+5>0; x+1>=0
\(C=\left|x+5\right|+\left|x+1\right|-x\)
=x+5+x+1-x
=x+6
d: D=|x+5|-2|2-x|
=|x+5|-2|x-2|
TH1: x<-5
=>x+5<0; x-2<0
\(D=|x+5|-2|x-2|\)
=-x-5-2(2-x)
=-x-5-4+2x
=x-9
TH2: -5<=x<2
\(D=|x+5|-2|x-2|\)
=x+5-2(2-x)
=x+5-4+2x
=3x+1
TH3: x>=2
\(D=|x+5|-2|x-2|\)
=x+5-2(x-2)
=x+5-2x+4
=-x+9
e: \(E=\left|x^2+3\right|+\left|x-1\right|-1\)
\(=x^2+3+\left|x-1\right|-1=\left|x-1\right|+x^2+2\)
TH1: x>=1
=>x-1>=0
=>\(E=x-1+x^2+2=x^2+x+1\)
TH2: x<1
=>x-1<0
=>\(E=1-x+x^2+2=x^2-x+3\)
f: \(F=\left|-x-4\right|-2\left|-x\right|-9\)
=|x+4|-2|x|-9
TH1: x<-4
=>x+4<0; x<0
=>F=-(x+4)-2(-x)-9
=-x-4+2x-9
=x-13
TH2: -4<=x<0
=>x+4>=0; x<0
=>F=(x+4)-2(-x)-9
=x-5+2x
=3x-5
TH3: x>=0
=>x+4>0; x>=0
=>F=x+4-2x-9
=-x-5
