Đáp án:
\(\dfrac{{2{x^3} + 4{x^2} - 3x - 8}}{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {{x^2} + 2x + 3} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 2; - 1;0;2} \right\}\\
A = \left( {\dfrac{{x + 2}}{{{x^2} + x}} + \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right).\dfrac{{{x^2} + 2x}}{{{x^2} + 2x + 3}}\\
= \left[ {\dfrac{{{{\left( {x + 2} \right)}^2}\left( {x - 2} \right) + \left( {x + 1} \right)\left( {{x^2} + x} \right)}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}} \right].\dfrac{{x\left( {x + 2} \right)}}{{{x^2} + 2x + 3}}\\
= \dfrac{{\left( {{x^2} + 4x + 4} \right)\left( {x - 2} \right) + {x^3} + {x^2} + {x^2} + x}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x\left( {x + 2} \right)}}{{{x^2} + 2x + 3}}\\
= \dfrac{{{x^3} - 2{x^2} + 4{x^2} - 8x + 4x - 8 + {x^3} + {x^2} + {x^2} + x}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x\left( {x + 2} \right)}}{{{x^2} + 2x + 3}}\\
= \dfrac{{2{x^3} + 4{x^2} - 3x - 8}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x\left( {x + 2} \right)}}{{{x^2} + 2x + 3}}\\
= \dfrac{{2{x^3} + 4{x^2} - 3x - 8}}{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {{x^2} + 2x + 3} \right)}}\\
b)Thay:x = 2\left( {KTM} \right)
\end{array}\)
( bạn xem lại đề có ghi nhầm dấu hay số không nhé )
