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 \(x=\frac{-3}{11}\)

a) \(1+\sqrt{3x+1}=3x\)

ĐKXĐ: \(3x+1\ge0\Leftrightarrow x\ge\frac{-1}{3}\)

\(\Rightarrow\sqrt{3x+1}=3x-1\)

\(\Leftrightarrow\left(\sqrt{3x+1}\right)^2=\left(3x-1\right)^2\)

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\)

\(\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{0;1\right\}\)

b) \(\sqrt{\frac{5x+7}{x+3}}=4\)

ĐKXĐ: \(\hept{\begin{cases}\frac{5x+7}{x+3}>0\\x+3\ne0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x>\frac{-7}{5}\\x< -3\end{cases}}\)

\(\Rightarrow\left(\sqrt{\frac{5x+7}{x+3}}\right)^2=4^2\)

\(\Leftrightarrow\frac{5x+7}{x+3}=16\)

\(\Leftrightarrow5x+7=16\left(x+3\right)\)

\(\Leftrightarrow5x+7=16x+48\)

\(\Leftrightarrow-11x=41\)

\(\Leftrightarrow x=\frac{-41}{11}\)(tm)

Vậy phương trình có 1 nghiệm duy nhất là \(x=\frac{-41}{11}\)