Respuesta :

Answer:

[tex]a=11, b=2[/tex]

Step-by-step explanation:

Keep in mind that [tex]4-\sqrt{5}[/tex] and [tex]4+\sqrt{5}[/tex] are conjugates of each other. While it's typically known for rationalization of the denominator, getting an integer answer helps to simplify the fraction as shown in steps 5-7:

[tex]\frac{(4-\sqrt{5})(4+\sqrt{5})}{2\sqrt{11}}[/tex]

[tex]\frac{4(4)+4(\sqrt{5})-4(\sqrt{5})-\sqrt{5}(\sqrt{5})}{2\sqrt{11}}[/tex]

[tex]\frac{16+4\sqrt{5}-4\sqrt{5}-5}{2\sqrt{11}}[/tex]

[tex]\frac{16-5}{2\sqrt{11}}[/tex]

[tex]\frac{11}{2\sqrt{11}}[/tex]

[tex]\frac{\sqrt{11}\sqrt{11}}{2\sqrt{11}}[/tex]

[tex]\frac{\sqrt{11}}{2}[/tex]

Therefore, [tex]a=11[/tex] and [tex]b=2[/tex]

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