Text Solve the following equation: 2xx+1+x−1x−2=x2+2x−1(x+1)(x−2).\frac{2x}{x+1} +\frac{x-1}{x-2} = \frac{x^2+2x-1}{(x+1)(x-2)}.x+12x+x−2x−1=(x+1)(x−2)x2+2x−1. Solution Existence conditions: x≠−1x\neq -1x=−1 and x≠2x\neq 2x=2. Multiply by (x+1)(x−2)(x+1)(x-2)(x+1)(x−2): 2x(x−2)+(x−1)(x+1)=x2+2x−1.2x(x-2) + (x-1)(x+1) = x^2+2x-1.2x(x−2)+(x−1)(x+1)=x2+2x−1. On the left: 2x2−4x+x2−1=3x2−4x−12x^2 - 4x + x^2 - 1 = 3x^2 - 4x - 12x2−4x+x2−1=3x2−4x−1. Hence 3x2−4x−1=x2+2x−1 ⇒ 2x2−6x=0 ⇒ 2x(x−3)=0.3x^2 - 4x - 1 = x^2 + 2x - 1 \;\Rightarrow\; 2x^2 - 6x = 0 \;\Rightarrow\; 2x(x-3) = 0.3x2−4x−1=x2+2x−1⇒2x2−6x=0⇒2x(x−3)=0. So x=0x=0x=0 or x=3x=3x=3, both acceptable. x=0 ∨ x=3\boxed{x=0 \ \vee\ x=3}x=0 ∨ x=3