Ta có: \(\frac{7x+4}{5}-x=\frac{\left|2x-5\right|}{2}\)
\(\Leftrightarrow\frac{2\left(7x+4\right)}{10}-\frac{10x}{10}=\frac{5\left|2x-5\right|}{10}\)
Suy ra: \(2\left(7x+4\right)-10x=5\left|2x-5\right|\)
\(\Leftrightarrow14x+8-10x=5\left|2x-5\right|\)
\(\Leftrightarrow5\left|2x-5\right|=4x+8\)(*)
Trường hợp 1: \(2x-5\ge0\Leftrightarrow2x\ge5\Leftrightarrow x\ge\frac{5}{2}\)
(*)\(\Leftrightarrow5\left(2x-5\right)=4x+8\)
\(\Leftrightarrow10x-25-4x-8=0\)
\(\Leftrightarrow6x-33=0\)
\(\Leftrightarrow6x=33\)
hay \(x=\frac{11}{2}\)(tm)
Trường hợp 2: \(2x-5< 0\Leftrightarrow2x< 5\Leftrightarrow x< \frac{5}{2}\)
(*)\(\Leftrightarrow5\left(5-2x\right)=4x+8\)
\(\Leftrightarrow25-10x-4x-8=0\)
\(\Leftrightarrow-14x-17=0\)
\(\Leftrightarrow-14x=17\)
hay \(x=\frac{-17}{14}\)(tm)
Vậy: \(S=\left\{\frac{11}{2};\frac{-17}{14}\right\}\)
\(\frac{7x+4}{5}-x=\frac{\left|2x-5\right|}{2}\)
\(\Leftrightarrow\frac{7x+4}{5}-\frac{5x}{5}=\frac{\left|2x-5\right|}{2}\)
\(\Leftrightarrow\frac{2x+4}{5}=\frac{\left|2x-5\right|}{2}\)
\(\Leftrightarrow2\left(2x+4\right)=5.\left|2x-5\right|\)
\(\Leftrightarrow4x+8=5\left|2x-5\right|\)
\(\Leftrightarrow\frac{4x+8}{5}=\left|2x-5\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=\frac{4x+8}{5}\\2x-5=\frac{-4x-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5\left(2x-5\right)=4x+8\\5\left(2x-5\right)=-4x-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}10x-25=4x+8\\10x-25=-4x-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}10x-4x=8+25\\10x+4x=-8+25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=33\\14x=17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{11}{2}\\x=\frac{17}{14}\end{matrix}\right.\)
Vậy...
