\(\dfrac{x+1}{x+2}+\dfrac{x-1}{x-2}=\dfrac{2x+1}{x+1}\) ĐKXĐ : x ≠ -2 ; x ≠ 2 ; x ≠ -1
\(\dfrac{\left(x+1\right)^2\left(x-2\right)+\left(x-1\right)\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}\)
(\(x^2\) + 2x +1)(x - 2) + (\(x^2\) - 1)(x + 2) = (2x + 1)(\(x^2\) - 4)
\(x^3\) + 2\(x^2\) + x - 2\(x^2\) - 4x -2 + \(x^3\) + 2\(x^2\) - x - 2 = 2\(x^3\) - 8x + \(x^2\) - 4
2\(x^3\) + 2\(x^2\) - 4x - 4 = 2\(x^3\) - 8x + \(x^2\) - 4
2\(x^3\) + 2\(x^2\) - 4x - 4 - 2\(x^3\) + 8x - \(x^2\) + 4 = 0
\(x^2\) + 4x = 0
x(x + 4) = 0
TH1 : x =0 (t/m) TH2 : x + 4 = 0
x = -4 (t/m)
Vậy x ∈ \(\left\{0;-4\right\}\)
