Đáp án:
\(\begin{array}{l}
2)\,\dfrac{1}{3}\\
3)\,x = 0;x = - 2;x = 2\\
4)\,a = 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\, \Leftrightarrow \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right):\left( {{x^2} + xy + {y^2}} \right)\\
= x - y\\
Thay\,x = \dfrac{2}{3};y = \dfrac{1}{3}\,ta\,duoc\\
x - y = \dfrac{2}{3} - \dfrac{1}{3} = \dfrac{1}{3}\\
3)\,3x\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow 3x\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = - 2
\end{array} \right.\\
4)\,{x^3} + {x^2} - x + a\\
= {x^3} + 2{x^2} - {x^2} - 2x + x + 2 + a - 2\\
= {x^2}\left( {x + 2} \right) - x\left( {x + 2} \right) + \left( {x + 2} \right) + a - 2\\
= \left( {x + 2} \right)\left( {{x^2} - x + 1} \right) + a - 2\\
gt \Rightarrow a - 2 = 0 \Leftrightarrow a = 2\\
5)\,P = {\left( {x + 2y} \right)^2} + 5\\
Vi\,{\left( {x + 2y} \right)^2} \ge 0\\
\Rightarrow {\left( {x + 2y} \right)^2} + 5 \ge 5 > 0\\
\Rightarrow P > 0\,voi\,moi\,x;y
\end{array}\)
