\(a,x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)

\(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)

\(x=\frac{2}{5}+\frac{1}{3}-\frac{1}{3}\)

\(x=\frac{2}{5}\)

Ý f hình như bạn viết thiếu đầu bài thì phải, bạn kiểm tra lại đầu bài nhé ! 

\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

=>\(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

=> \(x+\frac{3}{4}=\frac{1}{2}\)hoặc \(x+\frac{3}{4}=-\frac{1}{2}\).

 \(x+\frac{3}{4}=\frac{1}{2}\)

\(x=\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

\(x+\frac{3}{4}=-\frac{1}{2}\)

\(x=-\frac{1}{2}-\frac{3}{4}\)

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

=> \(x\)thuộc \(\left\{-\frac{1}{4};-\frac{5}{4}\right\}\)

vvvvv

Bạn đang trả lời câu hỏi của tớ hay là muốn chửi tớ?

\(\frac{1}{2}+\frac{1}{2}\div x=\frac{1}{3}\)

\(\Leftrightarrow\)\(\frac{1}{2}\div x=\frac{1}{3}-\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{2}\div x=\left(-\frac{1}{6}\right)\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\div\left(-\frac{1}{6}\right)\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\times\left(-6\right)\)

\(\Leftrightarrow\)\(x=-\frac{6}{2}=-3\)

Ta xét:

\(\left|3x+2\right|\ge0\)

\(\left|2x+5\right|\ge0\)

\(\left|x+1\right|\ge0\)

\(\Rightarrow\left|3x+2\right|+\left|2x+5\right|+\left|x+1\right|\ge0\)

\(\Rightarrow10x\ge0\)

\(\Rightarrow x\ge0\)

Do vậy: \(3x+2+2x+5+x+1=10x\)

\(\Rightarrow x\left(3+2+1\right)+\left(2+5+1\right)=10x\)

\(\Rightarrow6x+8=10x\)

\(\Rightarrow10x-6x=8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)