a: ĐKXĐ của A là \(x\ne-3\)
ĐKXĐ của B là \(x\notin\left\{1;-1\right\}\)
b: Khi x=3 thì \(A=\dfrac{3^2}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
c: \(B=\dfrac{3x}{x-1}-\dfrac{2x}{x+1}+\dfrac{x-3}{1-x^2}\)
\(=\dfrac{3x}{x-1}-\dfrac{2x}{x+1}-\dfrac{x-3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x\left(x+1\right)-2x\left(x-1\right)-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x^2+3x-2x^2+2x-x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+4x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+3}{x-1}\)
d: \(Q=A\cdot B=\dfrac{x+3}{x-1}\cdot\dfrac{x^2}{x+3}=\dfrac{x^2}{x-1}\)
Đặt Q=4
=>\(\dfrac{x^2}{x-1}=4\)
=>\(x^2=4x-4\)
=>\(x^2-4x+4=0\)
=>\(\left(x-2\right)^2=0\)
=>x-2=0
=>x=2(nhận)
