Bài 1:
a: \(\left|3x-5\right|-\frac52=-1\frac12\)
=>\(\left|3x-5\right|=-\frac32+\frac52=\frac22=1\)
=>\(\left[\begin{array}{l}3x-5=1\\ 3x-5=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=6\\ 3x=4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=\frac43\end{array}\right.\)
b: \(-2\left|\frac13-\frac{x}{2}\right|+7=4,5\)
=>\(-2\left|\frac12x-\frac13\right|=4,5-7=-2,5\)
=>\(\left|\frac12x-\frac13\right|=-2,5:\left(-2\right)=\frac{2.5}{2}=1,25=\frac54\)
=>\(\left[\begin{array}{l}\frac12x-\frac13=\frac54\\ \frac12x-\frac13=-\frac54\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac12x=\frac13+\frac54=\frac{4+15}{12}=\frac{19}{12}\\ \frac12x=-\frac54+\frac13=\frac{-15+4}{12}=-\frac{11}{12}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{19}{12}\cdot2=\frac{19}{6}\\ x=-\frac{11}{12}\cdot6=-\frac{11}{6}\end{array}\right.\)
c: |x-1|+3x=1(1)
TH1: x>=1
(1) sẽ trở thành: x-1+3x=1
=>4x=2
=>\(x=\frac12\) (loại)
TH2: x<1
(1) sẽ trở thành: 1-x+3x=1
=>2x=0
=>x=0(nhận)
d: |5x-3|-x=7(2)
TH1: \(x\ge\frac35\)
=>5x-3>=0
(2) sẽ trở thành: 5x-3-x=7
=>4x=10
=>\(x=\frac{10}{4}=\frac52\) (nhận0
TH2: x<3/5
=>5x-3<0
(2) sẽ trở thành: 3-5x-x=7
=>3-6x=7
=>6x=3-7=-4
=>\(x=-\frac46=-\frac23\) (nhận)
e: |2x+1|=|3x-2|
=>\(\left[\begin{array}{l}3x-2=2x+1\\ 3x-2=-2x-1\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-2x=2+1\\ 3x+2x=-1+2\end{array}\right.\)
=>\(\left[\begin{array}{l}x=3\\ 5x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac15\end{array}\right.\)
f: |4x-1|-2|2x+1|=0
=>|4x-1|=|4x+2|
=>4x-1=4x+2 hoặc 1-4x=4x+2
=>-1=2(loại) hoặc -8x=1
=>\(x=-\frac18\)
h: \(\left|x^2-4\right|+\left(-\frac12\right)^2=\frac14-\left|x+2\right|\)
=>\(\left|x^2-4\right|+\left|x+2\right|=0\)
=>|x+2|(|x-2|+1)=0
=>|x+2|=0
=>x+2=0
=>x=-2
i: |\(\left|2x-1\right|+\frac12\) |=4/5
=>\(\left[\begin{array}{l}\left|2x-1\right|+\frac12=\frac45\\ \left|2x-1\right|+\frac12=-\frac45\end{array}\right.\Rightarrow\left[\begin{array}{l}\left|2x-1\right|=\frac45-\frac12=\frac{8-5}{10}=\frac{3}{10}\\ \left|2x-1\right|=-\frac45-\frac12<0\left(loại\right)\end{array}\right.\)
=>\(\left[\begin{array}{l}2x-1=\frac{3}{10}\\ 2x-1=-\frac{3}{10}\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{3}{10}+1=\frac{13}{10}\\ 2x=-\frac{3}{10}+1=\frac{7}{10}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{13}{20}\\ x=\frac{7}{20}\end{array}\right.\)
Bài 2:
a: \(\left|2x-6\right|+\left|x+3\right|=8\) (1)
Th1; x<-3
=>x+3<0; 2x-6<0
(1) sẽ trở thành: -x-3+6-2x=8
=>-3x+3=8
=>-3x=5
=>x=-5/3(loại)
TH2: -3<=x<3
=>x+3>=0; 2x-6<0
(1) sẽ trở thành: x+3+6-2x=8
=>9-x=8
=>x=1(nhận)
TH3; x>=3
=>x+3>0; 2x-6>=0
(1) sẽ trở thành: x+3+2x-6=8
=>3x-3=8
=>3x=11
=>x=11/3(nhận)
b: |x+5|-|1-2x|=x
=>|x+5|-|2x-1|=x(2)
TH1: x<-5
(2) sẽ trở thành: -x-5-(1-2x)=x
=>-x-5+2x-1=x
=>x-6=x
=>-6=0(vô lý)
TH2: -5<=x<1/2
(2) sẽ trở thành: x+5-(1-2x)=x
=>x+5-1+2x=x
=>3x-4=x
=>2x=4
=>x=2(loại)
TH3: x>=1/2
(2) sẽ trở thành: x+5-(2x-1)=x
=>x+5-2x+1=x
=>-x+6=x
=>-2x=-6
=>x=3(nhận)
c: |\(x^2+2\left|x-\frac12\right|\) |=\(x^2+2\)
=>\(x^2+2\left|x-\frac12\right|=x^2+2\)
=>\(2\left|x-\frac12\right|=2\)
=>\(\left|x-\frac12\right|=1\)
=>\(\left[\begin{array}{l}x-\frac12=1\\ x-\frac12=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\\ x=-\frac12\end{array}\right.\)
d: \(\left|x+\frac12\right|+\left|x+\frac14\right|+\left|x+1\right|=x\)
=>x>=0
Phương trình sẽ trở thành: \(x=x+\frac12+x+\frac14+x+1=3x+\frac74\)
=>\(-2x=\frac74\)
=>\(x=-\frac78\) (loại)
