Bài 3:
a: \(\left|x+\frac14\right|-\frac34=5\)
=>\(\left|x+\frac14\right|=5+\frac34=\frac{23}{4}\)
=>\(\left[\begin{array}{l}x+\frac14=\frac{23}{4}\\ x+\frac14=-\frac{23}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{23}{4}-\frac14=\frac{22}{4}=\frac{11}{2}\\ x=-\frac{23}{4}-\frac14=-\frac{24}{4}=-6\end{array}\right.\)
b: \(2-\left|\frac32x-\frac14\right|=\frac54\)
=>\(\left|\frac32x-\frac14\right|=2-\frac54=\frac34\)
=>\(\left[\begin{array}{l}\frac32x-\frac14=\frac54\\ \frac32x-\frac14=-\frac54\end{array}\right.\Longrightarrow\left[\begin{array}{l}\frac32x=\frac54+\frac14=\frac64=\frac32\\ \frac32x=-\frac54+\frac14=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-1:\frac32=-\frac23\end{array}\right.\)
Bài 2:
a: A=|x-2,2|+|x-1,8|
=|x-1,8|+|x-2,2|
TH1: x<1,8
=>x-1,8<0; x-2,2<0
\(A=|x-1,8|+|x-2,2|\)
=1,8-x+2,2-x
=4-2x
TH2: 1,8<=x<2,2
=>x-1,8>=0; x-2,2<0
\(A=|x-1,8|+|x-2,2|\)
=x-1,8+2,2-x
=0,4
TH3: x>=2,2
=>x-1,8>0; x-2,2>=0
\(A=|x-1,8|+|x-2,2|\)
=x-1,8+x-2,2
=2x-4
b: \(B=\left|-x-1,4\right|+\left|x-2,6\right|\)
=>\(B=\left|x+1,4\right|+\left|x-2,6\right|\)
TH1: x<-1,4
=>x+1,4<0; x-2,6<0
\(B=\left|x+1,4\right|+\left|x-2,6\right|\)
=-x-1,4-x+2,6
=-2x+1,2
TH2: -1,4<=x<2,6
=>x+1,4>=0; x-2,6<0
\(B=\left|x+1,4\right|+\left|x-2,6\right|\)
=x+1,4+2,6-x
=4
TH3: x>=2,6
=>x+1,4>0; x-2,6>=0
\(B=\left|x+1,4\right|+\left|x-2,6\right|\)
=x+1,4+x-2,6
=2x-1,2
