\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
=> \(\dfrac{3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)}{15}\) \(=\dfrac{7x^2-14x-5}{15}\)
=> \(\dfrac{7x^2+22x-2}{15}=\dfrac{7x^2-14x-5}{15}\)
=> \(\dfrac{7x^2+22x-2-7x^2+14x+5}{15}=0\)
=>\(\dfrac{36x+3}{15}=0\)
=> 36x=-3
=>\(x=\dfrac{-1}{12}\)
Vay \(S=\left\{\dfrac{-1}{12}\right\}\)
Phương trình tương đương: \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\\ \Leftrightarrow7x^2+22x+3=7x^2-14x\\ \Leftrightarrow22x+3=-14x\\ \Leftrightarrow22x+14x=-3\\ \Leftrightarrow36x=-3\\ \Leftrightarrow x=-\dfrac{1}{12}\)
Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)
