Question

whole solution ???????

Answer 1

is it ryt????
..............

Answer 2

Answer: The value of a = 0 and b = 2

Step-by-step explanation:

Since we have given that

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=a+\sqrt{35}b$

We need to find the values of a and b.

First we rationalize the first part:

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\\\=\frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7})^2-(\sqrt{5})^2}\\\\=\frac{7+5+2\sqrt{35}}{7-5}\\\\=\frac{12+2\sqrt{35}}{2}\\\\=6+\sqrt{35}$

Similarly,

$\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}\\\\=\frac{(\sqrt{7}-\sqrt{5})^2}{(\sqrt{7})^2-(\sqrt{5})^2}\\\\=\frac{7+5-2\sqrt{35}}{7-5}\\\\=\frac{12-2\sqrt{35}}{2}\\\\=6-\sqrt{35}$

Now, put the both expressions together :

$6+\sqrt{35}-(6-\sqrt{35})=a+\sqrt{35}b\\\\0+2\sqrt{35}=a+\sqrt{35}b$

Comparing the coefficient of √35 and constant term:

We get that a = 0, b = 2