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

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

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

+) \(x+\frac{4}{15}=\frac{8}{5}\)

=> \(x=\frac{8}{5}-\frac{4}{15}=\frac{24}{15}-\frac{4}{15}=\frac{20}{15}=\frac{4}{3}\)

+) \(x+\frac{4}{15}=-\frac{8}{5}\)

=> \(x=-\frac{8}{5}-\frac{4}{15}\)

=> \(x=-\frac{24}{15}-\frac{4}{15}=-\frac{28}{15}\)

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

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

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

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

Ta có : \(|x+\frac{4}{15}|\ge0\forall x\)

\(\Rightarrow|x+\frac{4}{15}|=x+\frac{4}{15}\)

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

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

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

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

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

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

\(< =>\)\(|x+\frac{4}{15}|=1,6=\frac{8}{5}\)

\(< =>\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)

\(\Rightarrow\left(\left|x\right|-5\right)=0;\left(x3-8\right)=0;\left|x-7\right|=0\)

\(\left|x\right|-5=0\)

\(\left|x\right|=5\Rightarrow x=5và-5\)

\(x3-8=0\)

\(3x=8\)

\(x=\frac{8}{3}\)

\(\left|x-7\right|=0\)

\(x-7=0\)

\(x=7\)

bạn làm sai rồi

(IxI - 5)(x3 - 8).Ix - 7I = 0 =>(IxI - 5)=0 hc (x3 - 8)=0  hc  Ix - 7I = 0

                                    => IxI= 5   ;   x3 =8  ; x - 7 = 0 hc 7-X=0

                                   =>x=5  hc x=-5  ;  x=8/3  ;  x=7 

a) Vì \(\hept{\begin{cases}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{cases}}\)mà \(\left|x\right|+\left|x+2\right|=0\)nên \(\hept{\begin{cases}\left|x\right|=0\\\left|x+2\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}\)(vô lý)

[x]+[x+2]=0

=>[x]=0        =>x=0

    [x+2]=0

b)

\(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\)

Điều kiện \(x\ge0\)nên 

\(x\left(x^2-\frac{5}{4}\right)=x\)

\(\Leftrightarrow x^2-\frac{5}{4}=x:x=1\)

\(\Leftrightarrow x^2=\frac{9}{4}\)

\(\Rightarrow x=\sqrt{\frac{9}{4}}=\frac{3}{2}\)

xong từ mấy đời rồiNguyễn Vũ Dũng ak