` a) ` ` 2x(x - 1) - (x - 1)^2 = (x - 1)(x + 1) `
` => 2x(x - 1) - (x - 1)^2 - (x - 1)(x + 1) = 0 `
` => (x - 1)(2x - x + 1 - x - 1) = 0 `
` => 0.(x - 1) = 0 ` (luôn đúng $∀x$)
Vậy ` x in R `
` b) ` ` x^2 - 2x = -1 `
` => x^2 - 2x + 1 = 0 `
` => (x - 1)^2 = 0^2 `
` => x - 1 = 0 `
` => x = 1 `
Vậy ` x = 1 `