a: \(P=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}-\dfrac{x}{x+1}\right):\dfrac{x-1+x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}-\dfrac{x}{x+1}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}\)
\(=\dfrac{x^2+2x+1-x^2+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}=\dfrac{3x+1}{2x}\)
b: 1/P-x+1/8>=1
=>\(\dfrac{2x}{3x+1}-\dfrac{x+1}{8}>=1\)
=>\(\dfrac{16x-3x^2-4x-1-24x-8}{8\left(3x+1\right)}>=0\)
=>\(\dfrac{-3x^2-24x-9}{8\left(3x+1\right)}>=0\)
=>\(\dfrac{x^2+8x+3}{3x+1}< =0\)
TH1: x^2+8x+3<=0 và 3x+1>0
=>x>-1/3 và \(-4-\sqrt{13}< =x< =-4+\sqrt{13}\)
=>Loại
TH2: x^2+8x+3>=0 và 3x+1<0
=>x<-1/3 và (x<=-4-căn 13 hoặc x>=-4+căn 13)
=>x<=-4-căn 13
