\(\dfrac{x+1}{3x-4}=\dfrac{3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=3\left(3x-4\right)\)
\(\Leftrightarrow5x+5=9x-12\)
\(\Leftrightarrow5x-9x=-12-5\)
\(\Leftrightarrow-4x=-17\)
\(\Leftrightarrow x=\dfrac{17}{4}\)
\(ĐKXĐ:x\ne\dfrac{4}{3}\)
\(\dfrac{x+1}{3x-4}=\dfrac{3}{5}\Leftrightarrow3\left(3x-4\right)=5\left(x+1\right)\\ \Leftrightarrow9x-12=5x+5\Leftrightarrow4x=17\Rightarrow x=\dfrac{17}{4}\)
Ta có: \(\dfrac{x+1}{3x-4}=\dfrac{3}{5}\)
\(\left(x+1\right).5=\left(3x-4\right).3\)
\(5x+5=9x-12\)
\(5x-9x=-12-5\)
\(-4x=-7\)
\(x=\dfrac{-7}{-4}=\dfrac{7}{4}\)
\(\dfrac{x+1}{3x-4}=\dfrac{3}{5}\\ \Leftrightarrow\left(x+1\right).5=3\left(3x-4\right)\\ \Rightarrow5x+5=9x-12\\ \Rightarrow5x-9x=-12-5\\ \Rightarrow-4x=-17\\ \Rightarrow x=\dfrac{-17}{-4}=\dfrac{17}{4}\)
