Tìm x, biết: (a) x(x+1)−(x+2)(x−2)=0.

a) \(x\left( {x + 1} \right) - \left( {x + 2} \right)\left( {x - 2} \right) = 0\)

\({x^2} + x - {x^2} + 4 = 0\)

\(x + 4 = 0\)

\(x = - 4\)

Vậy \(x = - 4\).

b) \({(2x + 1)^2} - 4{(x + 1)^2} = 0\)

\(4{x^2} + 4x + 1 - 4({x^2} + 2x + 1) = 0\)

\(4{x^2} + 4x + 1 - 4{x^2} - 8x - 4 = 0\)

\( - 4x - 3 = 0\)

\( - 4x = 3\)

\(x = \frac{{ - 3}}{4}\)

Vậy \(x = \frac{{ - 3}}{4}\).

c) \({x^3} - 3{x^2} + 3x - 1 = 27\)

\({\left( {x - 1} \right)^3} = {3^3}\)

\(x - 1 = 3\)

\(x = 4\)

Vậy \(x = 4\).

d) \({(x + 2)^3} - \left( {{x^3} + 8} \right) = 0\)

\({x^3} + 6{x^2} + 12x + 8 - {x^3} - 8 = 0\)

\(6{x^2} + 12x = 0\)

\(6x\left( {x + 2} \right) = 0\)

Suy ra: \(x = 0\) hoặc \(x = - 2\)

Vậy \(x \in \left\{ {0; - 2} \right\}\).