Question

[Lê Tài Bảo Châu]

g) \(|9-7x|=5x-3\)

Vì \(|9-7x|\ge0;\forall x\)

\(\Rightarrow5x-3\ge0\)

\(\Rightarrow x\ge\frac{3}{5}\)

Ta có: \(|9-7x|=5x-3\)

\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=3-5x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-3-9\\-7x+5x=3-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1>\frac{3}{5}\left(chon\right)\\x=3>\frac{3}{5}\left(chon\right)\end{cases}}\)

Vậy \(x\in\left\{1;3\right\}\)

h) \(8x-|4x+1|=x+2\)

\(\Leftrightarrow|4x+1|=7x+2\)

Vì \(|4x+1|\ge0;\forall x\)

\(\Rightarrow7x+2\ge0\)

\(\Rightarrow x\ge\frac{-2}{7}\)

Ta có: \(|4x+1|=7x+2\)

\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}< \frac{-2}{7}\left(loai\right)\\x=\frac{-3}{11}>\frac{-2}{7}\left(chon\right)\end{cases}}\)

Vậy \(x=\frac{-3}{11}\)