Đáp án:
${x^2} + xy - {y^2} = \left( {x + y\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right)} \right)\left( {x + y\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)} \right)$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} + xy - {y^2}\\
= {x^2} + 2.x.\dfrac{y}{2} + {\left( {\dfrac{y}{2}} \right)^2} - {\left( {\dfrac{y}{2}} \right)^2} - {y^2}\\
= {\left( {x + \dfrac{y}{2}} \right)^2} - \dfrac{{5{y^2}}}{4}\\
= {\left( {x + \dfrac{y}{2}} \right)^2} - {\left( {\dfrac{{y\sqrt 5 }}{2}} \right)^2}\\
= \left( {x + \dfrac{y}{2} - \dfrac{{y\sqrt 5 }}{2}} \right)\left( {x + \dfrac{y}{2} + \dfrac{{y\sqrt 5 }}{2}} \right)\\
= \left( {x + y\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right)} \right)\left( {x + y\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)} \right)
\end{array}$
Vậy ${x^2} + xy - {y^2} = \left( {x + y\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right)} \right)\left( {x + y\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)} \right)$
