\(a,\left(x+1\right)^2=x^2-2x-3\)
\(\Leftrightarrow\left(x+1\right)^2=x^2-3x+x-3\)
\(\Leftrightarrow\left(x+1\right)^2=x\left(x-3\right)+\left(x-3\right)\)
\(\Leftrightarrow\left(x+1\right)^2=\left(x+1\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-x+3\right)=0\)
\(\Leftrightarrow4\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy pt có tập nghiệm S = { - 1 }
b, ĐKXĐ :\(\left\{{}\begin{matrix}1-2x\ne0\\1+2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
\(\dfrac{3}{1-2x}=\dfrac{2}{1+2x}-\dfrac{3x+12}{1-4x^2}\)
\(\Leftrightarrow\dfrac{3\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}=\dfrac{2\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}-\dfrac{3x+12}{\left(1-2x\right)\left(1+2x\right)}\)
\(\Rightarrow3+6x=2-4x-3x-12\)
\(\Leftrightarrow6x+4x+3x=2-12-3\)
\(\Leftrightarrow13x=-13\)
\(\Leftrightarrow x=-1\) ( t/m )
Vậy pt có tập nghiệm S = { - 1 }