Cho hai biểu thức :  \(A =...

D

- Hướng dẫn giải

Để A=B thì \( 1 + \frac{1}{{2 + x}} = \frac{{12}}{{{x^3} + 8}}\)

ĐKXĐ: x≠−2

\(\begin{array}{l} 1 + \frac{1}{{2 + x}} = \frac{{12}}{{{x^3} + 8}} \Leftrightarrow 1 + \frac{1}{{x + 2}} = \frac{{12}}{{(x + 2)({x^2} - 2x + 4)}} \Leftrightarrow \frac{{{x^3} + 8 + {x^2} - 2x + 4}}{{(x + 2)({x^2} - 2x + 4)}} = \frac{{12}}{{(x + 2)({x^2} - 2x + 4)}}\\ \to {x^3} + 8 + {x^2} - 2x + 4 = 12 \Leftrightarrow {x^3} + {x^2} - 2x = 0 \to \left[ \begin{array}{l} x = 0\\ {x^2} + x - 2 = 0 \end{array} \right. \to \left[ \begin{array}{l} x = 0\\ (x - 1)(x + 2) = 0 \end{array} \right. \to \left[ \begin{array}{l} x = 0(tm)\\ x = 1(tm)\\ x = - 2(ktm) \end{array} \right. \end{array}\)

Vậy để A=B thì x=0 hoặc x=1