\(b,x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\\ \Rightarrow x^2+x-\left(x^2+2x-3x-6\right)-7=0\\ \Rightarrow x^2+x-\left(x^2-x-6\right)-7=0\\ \Rightarrow x^2+x-x^2+x+6-7=0\\ \Rightarrow2x-1=0\\ \Rightarrow x=\dfrac{1}{2}\)
\(c,ĐKXĐ:x\ne\pm1\)
\(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2-3x-x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\\ \Rightarrow x^2-4x+3=x^2\\ \Rightarrow x^2-x^2+4x-3=0\\ \Leftrightarrow4x-3=0\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
