\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\frac{1}{12}\)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow\)12x2+12x+3-5x2+10x-5-7x2+14x+5=0
\(\Leftrightarrow\)36x+3=0
\(\Leftrightarrow\)36x=-3
\(\Leftrightarrow\)x=\(-\frac{1}{12}\)
Vậy pt có tập nghiêm S={\(-\frac{1}{12}\)}
