Question

pllzzz simplify this question also​

Answer 1

Given :-

$\sf{\implies \frac{\sqrt{5} -1 }{\sqrt{5} +1 } + \frac{\sqrt{5} + 1 }{\sqrt{5}-1 } = a + b\sqrt{5} }$

To Find :-

$\sf{\implies \; Value \; of \; A \; and \; B }$

Solution :-

$\sf{\implies \frac{\sqrt{5} -1 }{\sqrt{5} +1 } + \frac{\sqrt{5} + 1 }{\sqrt{5}-1 } }$

Rationalizing these two fractions seperately :-

$\sf{\implies \frac{\sqrt{5} -1 }{\sqrt{5} +1 } \times \frac{\sqrt{5} -1 }{\sqrt{5} -1 } + \frac{\sqrt{5} + 1 }{\sqrt{5}-1 } \times \frac{\sqrt{5} +1 }{\sqrt{5} +1 } }$

$\sf{\implies \frac{[\sqrt{5} -1 ]^{2} }{[\sqrt{5}]^{2} - 1 } + \frac{[\sqrt{5} + 1]^{2} }{[\sqrt{5}]^{2} -1 } }$

$\sf{\implies \frac{[\sqrt{5} -1 ]^{2} + [ \sqrt{5} +1 ] ^{2} }{[\sqrt{5}]^{2} - 1 } }$

$\sf{\implies \frac{[\sqrt{5} ]^{2} + 1 -2\sqrt{5} + [ \sqrt{5} ] ^{2} + 1 + 2\sqrt{5} }{5 - 1 } }$

$\sf{\implies \frac{5+1 - 2\sqrt{5} + 5 + 1 + 2\sqrt{5} }{4} }$

$\sf{\implies \frac{12 }{4} \rightarrow 3 }$

$\sf{\implies a + b\sqrt{5} = 3 + 0\sqrt{5} }$

Hence , A = 3 and B = 0

Answer 2

Explanation:

ԶŲƐʂɬıƠИ :-

$\sf \green{ \to \: \frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=a+b\sqrt{5}} \\ \\ </p><p>$

ƠИ ʂƠƖ۷ıИƓ :-

$</p><p> \sf\orange{\to\frac{(\sqrt{5}+1)^2+(\sqrt{5}-1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}=a+b\sqrt{5}} \\ \\ </p><p>$

$\sf \blue{\to\frac{(\sqrt{5}+1)^2+(\sqrt{5}-1)^2}{5-1}=a+b\sqrt{5} }$

$\sf\pink{ \to\frac{5+1+2\sqrt{5}+5+1-2\sqrt{5}}{4}=a+b\sqrt{5} } \\ \\ </p><p>$

$\sf \purple{ \to\frac{12+0\sqrt{5}}{4}=a+b\sqrt{5}}$

$\bf \sf \red{ \to \: 3+0\sqrt{5}=a+b\sqrt{5}} \\ \: \bf \blue{ \to a =3 \: nd \: b \: = 0 }$