{x^2} + 2x + 1 = (x + 2)\sqrt {{x^2} + 2} \Rightarrow {(x + 1)^4} = ({x^2} + 4x + 4)({x^2} + 2) \hfill \\ \Rightarrow {x^4} + 4{x^3} + 6{x^2} + 4x + 1 = {x^4} + 4{x^3} + 6{x^2} + 8x + 8 \hfill \\ \Rightarrow 4x = - 7 \to x = - \frac{7}{4} \hfill \\ \end{gathered} \][/quote]
$\
Leftrightarrow (x+1)^4=(x^2+4x+4)(x^2+2)
$$\
Lef
trightarrow x^4+4x^3+4x+1=x^4+4x^3+6x^2+8x+8
$$4x=-7\
Lef
trightarrow x=-\frac{7}{4}
$