chia khoang 

nghiệm của ba số hạng là 

x=3

x= -4/3 

x=-1/2

-4/3<-1/2<3

x<-4/3 

-(x-3)-(3x+4)=-(2x+1)

-x+3-3x-4=-2x-1=> 2x=0=> x=0 loại

-4/3<=x<-1/2

-(x-3)+3x+4=-2x-1

-x+3+3x+4=-2x-1=>4x=-7=>x=-7/4 loại

-1/2<=x<3

-x+3+3x+4=2x+1  2x+7=2x+1=>vô gnhiệm

x>=3

x-3+3x+4=2x+1

2x=0

x=0 loại

(1) vô nghiệm mỏi rồi 

a: =>|2x+3|=2+2x-5=2x-3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(2x-3-2x-3\right)\left(2x-3+2x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

b: Trường hợp 1: x<-3

Pt sẽ là -x-3+1-2x=10

=>-3x-2=10

=>-3x=8

hay x=-8/3(loại)

Trường hợp 2: -3<=x<1/2

Pt sẽ la x+3+1-2x=10

=>4-x=10

hay x=-6(loại)

Trường hợp 3: x>=1/2

Pt sẽ là x+3+2x-1=10

=>3x+2=10

hay x=8/3(nhận)

a, \(\left|2x-5\right|=4\)

\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}\)

b, \(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Rightarrow\left|2x-3\right|=\left|3x+2\right|\)

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)

c, \(\left|x+3\right|-\left|3x+2\right|=x+2\)

Ta có: x + 3 = 0 => x = -3

           3x + 2 = 0 => x = -2/3

Lập bảng xét dấu: 

Với x < -3

Ta có: -x - 3 + 3x + 2 = x + 2

<=> 2x - 1 = x + 2

<=> x = 3 ( ko t/mãn )

Với -3 ≤ x < -2/3

Ta có: x + 3 + 3x + 2 = x + 2

<=> 4x + 5 = x + 2

<=> 3x = -3

<=> x = -1 ( t/mãn )

Với -2/3 ≤ x 

Ta có: x + 3 - 3x - 2 = x + 2

<=> -2x + 1 = x + 2

<=> -3x = 1

<=> x = -1/3 ( t/mãn )

Vậy....

d, \(\left||x-1|-5\right|=x+5\)

Đk: x + 5 ≥ 0 => x ≥ -5

\(\Rightarrow\orbr{\begin{cases}\left|x-1\right|-5=x+5\\\left|x-1\right|-5=-x-5\end{cases}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=x+25\\\left|x-1\right|=-x\left(Loai\right)\end{cases}}}\)

Giải \(\left|x-1\right|=x+25\)

\(\Rightarrow\orbr{\begin{cases}x-1=-x-25\\x-1=x+25\end{cases}\Rightarrow\orbr{\begin{cases}2x=-24\\0x=26\left(Loai\right)\end{cases}\Rightarrow x}=-12}\)( ko t/mãn )

Vậy x \(\in\varnothing\)

a)x=1
b)x=1
tick cho mình nha

a,|x-3|+x=7

|x-3| =7-x

Th1: x-3 =7-x Th2: x-3=-7-x

x+x=7+3 x+x=-7+3

2x =10 2x =-4

x =10:2 x =-4:2

x =5 x =-2

a: =>|x-3|=7-x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =7\\\left(x-3-7+x\right)\left(x-3+7-x\right)=0\end{matrix}\right.\Leftrightarrow x=5\)

b: TH1: x<1/2

Pt sẽ là 3-x-(1-2x)=x

=>3-x-1+2x=x

=>x+2=x(loại)

TH2: 1/2<=x<3

Pt sẽ là x=3-x-(2x-1)=3-x-2x+1=-3x+4

=>4x=4

=>x=1(nhận)

TH3: x>=3

Pt sẽ là x-3-2x+1=x

=>x=-x-2

=>2x=-2

=>x=-1(loại)

\(a)\left|2x-5\right|=4\)\(\Rightarrow2x-5=\pm4\)

\(Với\)\(2x-5=4\Rightarrow2x=9\Rightarrow x=\frac{9}{2}\)

\(Với\)\(2x-5=-4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

\(Vậy\)\(x=\frac{9}{2};x=\frac{1}{2}\)

\(b)\left|2x-3\right|-\left|3x+2\right|=0\)

\(Vì\)\(\left|2x-3\right|\ge0;\left|3x+2\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}2x-3=0\\3x+2=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3\\3x=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=\frac{-2}{3}\end{cases}}}\)

\(Vậy\)\(x=\frac{3}{2};x=\frac{-2}{3}\)

11: |2x-3|-1/3=0

=>|2x-3|=1/3

=>\(\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{10}{3}\\2x=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

12: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)

=>\(\left|x+\dfrac{1}{4}\right|=\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}\)

=>\(\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{12}\end{matrix}\right.\)

13: \(\left|x-1\right|-2x=\dfrac{1}{2}\)

=>\(\left|x-1\right|=2x+\dfrac{1}{2}\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}\right)^2=\left(x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}-x+1\right)\left(2x+\dfrac{1}{2}+x-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

14: \(3x-\left|x+15\right|=\dfrac{5}{4}\)

=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16.25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\)

=>\(x=8.125\)

1: |1-5x|-1=3

=>|5x-1|=4

=>5x-1=4 hoặc 5x-1=-4

=>5x=5 hoặc 5x=-3

=>x=1 hoặc x=-3/5

2: 4|2x-1|+3=15

=>4|2x-1|=12

=>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=2 hoặc x=-1

3,\(\left|x+4\right|=2x+1\)

TH1: x+4≥0⇔x≥-4,pt có dạng:

x+4=2x+1⇔-x=-3⇔x=3(t/m)

TH2:x+4<0⇔x<-4,pt có dạng:

-x-4=2x+1⇔-3x=5⇔x=\(\dfrac{-5}{3}\)(loại)

Vậy pt đã cho có tập nghiệm S=\(\left\{3\right\}\)

4,\(\left|3x+4\right|=x-3\)

TH1: 3x-4≥0⇔3x≥4⇔x≥\(\dfrac{4}{3}\),pt có dạng:

3x-4=x-3⇔2x=1⇔x=\(\dfrac{1}{2}\)(loại)

TH2: 3x-4<0⇔3x<4⇔x<\(\dfrac{4}{3}\),pt có dạng:

-3x+4=x-3⇔-4x=-7  ⇔x=1,75(loại)

Vậy pt đã cho vô nghiệm