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Answer:
1
Step-by-step explanation:
Identity used (a - b)(a+b)= a² -b²
[tex]\text{Conjugate of $\sqrt{5}+\sqrt{4} = \sqrt{5}-\sqrt{4}$}[/tex]
[tex]\sf \dfrac{1}{\sqrt{5}+\sqrt{4}}=\dfrac{1*(\sqrt{5}-\sqrt{4})}{(\sqrt{5}+\sqrt{4})(\sqrt{5}-\sqrt{4})}[/tex]
[tex]\sf =\dfrac{\sqrt{5}-\sqrt{4}}{(\sqrt{5})^2-(\sqrt{4})^2}\\ \\ = \dfrac{\sqrt{5} -\sqrt{4}}{5-4}\\\\ = \dfrac{\sqrt{5}-\sqrt{2*2}}{1}\\\\= \sqrt{5}-2[/tex]
In the same way,
[tex]\dfrac{1}{\sqrt{5}+\sqrt{6}}=\sqrt{6}-\sqrt{5}\\\\\dfrac{1}{\sqrt{7}+\sqrt{6}}=\sqrt{7}-\sqrt{6}\\\\\dfrac{1}{\sqrt{8}+\sqrt{7}}=\sqrt{8}-\sqrt{7}\\\\\dfrac{1}{\sqrt{9}+\sqrt{8}}=\sqrt{9}-\sqrt{8}= \sqrt{3*3}-\sqrt{8}=3-\sqrt{8}[/tex]
[tex]\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{8}}+\dfrac{1}{\sqrt{8}+\sqrt{9}} \\\\\\=\sqrt{5}-2+\sqrt{6}-\sqrt{5} +\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+ 3 -\sqrt{8}[/tex]
= -2 + 3
= 1