Question

Pl solve question number 14.

Please don't say that the pic is not good
Fastttt

Happy new year

Answer 1

happy new year ,
here is ur answer ,
first rationalise them
so here u go
see attachment
a = 21/45 and b = 45/11
hope it helps

Answer 2

Answer:

a=0,b=1

Step-by-step explanation:

$Given Equation is \frac{7+\sqrt{5}}{7-\sqrt{5}} - \frac{7-\sqrt{5}}{7+\sqrt{5}} = a + \frac{7}{11}\sqrt{5}b$

$=>\frac{7+\sqrt{5}}{7-\sqrt{5}}*\frac{7+\sqrt{5}}{7+\sqrt{5}} - \frac{7-\sqrt{5}}{7+\sqrt{5}}*\frac{7-\sqrt{5}}{7-\sqrt{5}}=a+\frac{7}{11}\sqrt{5}$

$=>\frac{(7+\sqrt{5})^2}{(7)^2 - (\sqrt{5})^2} - \frac{(7-\sqrt{5})^2}{(7)^2-(\sqrt{5})^2}=a+\frac{7}{11}\sqrt{5}$

$=>\frac{49+5+14\sqrt{5}}{44}-\frac{49+5-14\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}$

$=>\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b$

$=>\frac{28\sqrt{5}}{44}=a+\frac{7}{11}\sqrt{5}b$

$=>\frac{7\sqrt{5}}{11}=a+\frac{7}{11}\sqrt{5}b$

$=>0+\frac{7}{11}\sqrt{5}(1)=a+\frac{7}{11}\sqrt{5}b$

$=>\boxed{a=0,b=1}$

Hope it helps!