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

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

=> |x + 4/15| = 59/10

=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{59}{10}\\x+\frac{4}{15}=-\frac{59}{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{169}{30}\\x=\frac{-185}{30}\end{cases}}\)

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

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

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

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

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

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

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

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

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

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

Còn lại bn tự làm nha 

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

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

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

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

|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

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

/x+4/15/-15/4=-\(\frac{43}{20}\)

x+4/15=\(\frac{43}{20}+\frac{15}{4}\)

x+4/15=59/10

    x=59/10-4/15

    x=\(5\frac{19}{30}\)

/x + \(\frac{4}{15}\)/ - 3,75 = -2,15

=> /x + \(\frac{4}{15}\)/ = 1,6

=> \(x+\frac{4}{5}=1,6\) hoặc \(x+\frac{4}{5}=-1,6\)

=> \(x=0,8\)hoặc \(x=-2,4\)

c) 22/5 + 51/9 + 11/4 + 3/5 + 1/3 + 1/4
= 22/5 +3/5 +51/9 + 1/3 +11/4+1/4
= (22/5 +3/5) +(51/9 + 3/9) +(11/4+1/4)
= 25/5 +54/9 +12/4
= 5 +6 +3
= 14
d) (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15) 
= (5/30 + 3/30 +2/30 ) :(5/30 +3/30 -2/30)
= 10/30 : 6/30
= 1/3 : 1/5
= 5/3

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

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

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

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

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

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

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

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

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

| x + \(\frac{4}{5}\)| - | - 3,75 | = - | - 2,15 |

=> | x + \(\frac{4}{5}\)| - 3,75 = - 2,15

=> | x + \(\frac{4}{5}\)| = 1,6

=> \(\orbr{\begin{cases}x+\frac{4}{5}=1,6\\x+\frac{4}{5}=-1,6\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=-\frac{12}{5}\end{cases}}\)

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

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

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

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

+) \(x+\frac{4}{15}=16\)

\(\Rightarrow x=\frac{236}{15}\)

+) \(x+\frac{4}{15}=-16\)

\(\Rightarrow x=-\frac{244}{15}\)

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

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

|x+4/15| = 1

TH1:

x+4/15 =1

x= 11/15

TH2:

x+ 4/15 = -1

x= -19/15

Vậy x=11/15 và x= -19/15

X= 4/3

Nhớ K mk nhe!!! 

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

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

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

\(\Rightarrow\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=\frac{-28}{15}\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=\frac{-28}{15}\end{array}\right.\)