Question

please answer correct it is my request ​

Answer 1

Step-by-step explanation:

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Answer 2

$\sf \dfrac{4+\sqrt{5}}{4-\sqrt{5}}= \Big[a+b\sqrt{5}\Big]$

We have to find the value of a and b

Then , take L.H.S

$\longmapsto\sf \dfrac{4+\sqrt{5}}{4-\sqrt{5}}\times \dfrac{4+\sqrt{5}}{4+\sqrt{5}}\\ \\ \\ \longmapsto\sf \dfrac{(4+\sqrt{5})^2}{(4)^2-(\sqrt{5})^2}$

$\small{\sf {\dag \ \ (a+b)^2= a^2+b^2+2ab}}\\ \\ \sf{\small{\dag\ \ a^2-b^2= (a+b)(a-b)}}$

$\longmapsto\sf \dfrac{(4)^2+(\sqrt{5})^2+2\times 4\times \sqrt{5}}{16-5}\\ \\ \\ \longmapsto\sf \dfrac{16+5+8\sqrt{5}}{11}\\ \\ \\ \longmapsto\sf \dfrac{21+8\sqrt{5}}{11}$

Now the value of a and b compare them !

$\longmapsto\sf \dfrac{21+8\sqrt{5}}{11}= a+b\sqrt{5}\\ \\ \\ \longmapsto\sf \dfrac{21}{11}+\dfrac{8\cancel{\sqrt{5}}}{11}= a+b\cancel{\sqrt{5}}\\ \\ \\ \longmapsto\sf \dfrac{21}{11} +\dfrac{8}{11}=a+b$

$\boxed{\sf{ a= \dfrac{21}{11}}}$

$\boxed{\sf{ b= \dfrac{8}{11}}}$