[sahaete]

โดยใช้วิธี Componendo et dividendo จะได้
$\qquad \dfrac{{\left( {\sqrt {x + 4} - \sqrt {x - 1} } \right) + \left( {\sqrt {x + 4} +
\sqrt {x - 1} } \right)}}{{\left( {\sqrt {x + 4} - \sqrt {x - 1} } \right) -
\left( {\sqrt {x + 4} + \sqrt {x - 1} } \right)}} = \dfrac{{\left( {x - 4} \right) + 5}}
{{\left( {x - 4} \right) - 5}}$

$\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \dfrac{{2\sqrt {x + 4} }}{{ - 2\sqrt {x - 1} }} = \dfrac{{x + 1}}{{x - 9}}$
$\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \dfrac{{x + 4}}{{x - 1}} = \dfrac{{{x^2} + 2x + 1}}{{{x^2} - 18x + 81}}$
$\qquad\quad \qquad \qquad \qquad \quad \qquad\dfrac{{\left( {x + 4} \right) + \left( {x - 1} \right)}}{{\left( {x + 4} \right) -
\left( {x - 1} \right)}} = \dfrac{{\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} - 18x + 81}
\right)}}{{\left( {{x^2} + 2x + 1} \right) - \left( {{x^2} - 18x + 81} \right)}}$

จะได้สมการดังนี้ $\quad \quad \dfrac{{2x + 3}}{5} = \dfrac{{2{x^2} - 16x + 82}}{{20x - 80}} = \dfrac{{{x^2} - 8x + 41}}{{10x - 40}}$
และเมื่อแก้สมการปกติ คำตอบ$\qquad \qquad x = 5$