a) \(\left|2,5-x\right|-1,3=0\)

th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)

th2: \(2,5-x< 0\Leftrightarrow x>2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)

vậy \(x=1,2;x=3,8\)

b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)

c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)

th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)

th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)

vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)

d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)

th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)

vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)

e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm

│x+14/15│-3,75=-2,15

│x+14/15│=-2,15+3,75=1,6

Suy ra :

TH1 :x+14/15=1,6

x=2/3

TH2 :x+14/15=-1,6

x=-38/15

\(\left|x+\dfrac{14}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|-3,75=-2,15\)

\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|=-2,15+3,75\)

\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{14}{15}=-1,6\\x+\dfrac{14}{15}=1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-38}{15}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy....

Chúc các bạn học tốt !!!

│x+14/15│-3,75=-2,15

│x+14/15│=-2,15+3,75=1,6

Suy ra :

TH1 :x+14/15=1,6

x=2/3

TH2 :x+14/15=-1,6

x=-38/15

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\frac{4}{15}\right|=1,6\)

Ta có 2 trường hợp

trường hợp 1

\(x+\frac{4}{15}=1,6\)

\(x=1,6-\frac{4}{15}\)

\(x=\frac{4}{3}\)

Trường hợp 2

\(x+\frac{4}{15}=-1,6\)

\(x=-1,6-\frac{4}{15}\)

\(x=\frac{-28}{15}\)

Vậy \(x=\left\{-\frac{28}{15};\frac{4}{3}\right\}\)

a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=1.6=\dfrac{8}{5}\)

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

c: =>x-y=0 và y+9/25=0

=>x=y=-9/25

d: =>-1/3<x-3/5<1/3

=>4/15<x<14/15

e: =>|x+5,5|>5,5

=>x+5,5>5,5 hoặc x+5,5<-5,5

=>x>0 hoặc x<-11

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=1,6\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{15}=1,6\\x+\frac{4}{15}=-1,6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}}\)

|x+4/15|-3,75=-2,15

|x+4/15|=1,6

+)x+4/15=1,6

x=4/3

+)x+4/15=-1,6

x=-28/15

làm tiếp cái trước(ấn nhầm)

\(x=\frac{5}{42}-\frac{15}{28}\) 

\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)

\(x=\frac{20}{168}-\frac{90}{168}\)

\(x=\frac{-70}{168}\)

\(x=\frac{-5}{12}\)

2. 

1.

 \(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)

\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)

\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)

/x+\(\frac{4}{15}\)/ \(-\)\(3,75\)\(=\)\(-2,15\)

/ \(x+\frac{4}{15}\)/ \(=\)\(-2,15+3,75\)

/ \(x+\frac{4}{15}\)/ \(=\)\(\frac{8}{5}\)

TH1: \(x+\frac{4}{15}=\frac{8}{5}\)                                          TH2: \(x+\frac{4}{15}=-\frac{8}{5}\)

\(=>x=\frac{8}{5}-\frac{4}{15}\)                                    \(=>x=-\frac{8}{5}-\frac{4}{15}\)

\(=>x=\frac{24}{15}-\frac{4}{15}\)                                   \(=>x=-\frac{24}{15}-\frac{4}{15}\)

\(=>x=\frac{20}{15}\)                                                 \(=>x=-\frac{28}{15}\)

\(=>x=\frac{4}{3}\)                                                         

\(\left|x+\frac{4}{15}\right|-\left|-3.75\right|=-\left|-2,15\right|\)

\(\Rightarrow\left|x+\frac{4}{15}\right|-\frac{15}{4}=-\frac{43}{20}\)

\(\Rightarrow\left|x+\frac{4}{15}\right|=-\frac{43}{20}+\frac{15}{4}\)

\(\Rightarrow\left|x+\frac{4}{15}\right|=\frac{8}{5}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\) 

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\) 

\(\left|x+\frac{4}{15}\right|=1,6\) 

=>  \(x+\frac{4}{15}=1,6\)  hoặc  \(x+\frac{4}{15}=-1,6\) 

=>  \(x=\frac{4}{3}\) hoặc \(x=\frac{-28}{15}\) 

Vậy..

=> x + 4/15 - 3,75 = -2,15

=> x +4/15 =-2,15 + 3,75

=> x+4/15 = 1,6

=> x = 1,6-4/15

=> x = 4/3