Ta có: \(\sqrt{\left(\dfrac{1}{3}+x\right)^2}:\left(x+\dfrac{3}{4}\right)=\dfrac{7}{9}\)
\(\Rightarrow\left|\dfrac{1}{3}+x\right|:\left(x+\dfrac{3}{4}\right)=\dfrac{7}{9}\)
\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{1}{3}+x\right):\left(x+\dfrac{3}{4}\right)=\dfrac{7}{9}\\\left(-\dfrac{1}{3}-x\right):\left(x+\dfrac{3}{4}\right)=\dfrac{7}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}+x=\dfrac{7}{9}x+\dfrac{7}{12}\\-\dfrac{1}{3}-x=\dfrac{7}{9}x+\dfrac{7}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{7}{9}x=\dfrac{7}{12}-\dfrac{1}{3}\\-x-\dfrac{7}{9}x=\dfrac{7}{12}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{9}x=\dfrac{1}{4}\\-\dfrac{16}{9}x=\dfrac{11}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-33}{64}\end{matrix}\right.\)
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