Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)^2=17\)
\(\Leftrightarrow4x^2+12x+9-4x^2+8x-4=17\)
\(\Leftrightarrow20x=12\)
hay \(x=\dfrac{3}{5}\)
\(\left(2x+3\right)^2-4\left(x-1\right)^2=17\)
\(\Leftrightarrow\left(2x+3-2x+2\right)\left(2x+3+2x-2\right)=17\)
\(\Leftrightarrow5\left(4x+1\right)=17\)
\(\Leftrightarrow20x+5=17\)
\(\Leftrightarrow x=\dfrac{3}{5}\)