<=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}-\frac{7x^2-14x-5}{15}=0\)
<=> \(\frac{12x^2+12x+3-5x^2+10x-5-7x^2+14x+5}{15}=0\)
=> 36x + 3 = 0
<=> 36x = -3
<=> x = -1/12
Vậy S = { -1/12 }
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{4x^2+4x+1}{5}-\frac{x^2-2x+1}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\Leftrightarrow x=-\frac{3}{36}=-\frac{1}{12}\)
