Find the values of a and b in each of the fol | vedclass

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(D) Given equation: $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$Rationalizing the denominators:$\frac{(7+\s

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$0, 1$

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(D) Given equation: $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$Rationalizing the denominators:$\frac{(7+\sqrt{5})^2}{(7-\sqrt{5})(7+\sqrt{5})} - \frac{(7-\sqrt{5})^2}{(7+\sqrt{5})(7-\sqrt{5})} = a+\frac{7}{11} \sqrt{5} b$Using $(a+b)^2 = a^2+b^2+2ab$ and $(a-b)^2 = a^2+b^2-2ab$:$\frac{49+5+14\sqrt{5}}{49-5} - \frac{49+5-14\sqrt{5}}{49-5} = a+\frac{7}{11} \sqrt{5} b$$\frac{54+14\sqrt{5}}{44} - \frac{54-14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$$\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$$\frac{28\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$Simplifying the fraction $\frac{28}{44}$ by dividing by $4$:$\frac{7\sqrt{5}}{11} = a+\frac{7}{11} \sqrt{5} b$Comparing both sides,we get $a=0$ and $b=1$.

Find the values of $a$ and $b$ in each of the following:
$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$

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Find the values of $a$ and $b$ in each of the following:$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$