Question

plzz help me to find both answer

Answer 1

2)- 

[x^(a-b)]^(a+b)  * [x^(b-c)]^(b+c)  *  [x^(c-a)]^(c+a)

by formula,  a^2 -b^2 = (a+b)(a-b)

[x^(a^2 -b^2)] * [x^(b^2 -c^2] * [x^(c^2 -a^2)]



base is same, powers will add

x^[a^2 -b^2 +b^2 -c^2 +c^2 -a^2]

= x^(0)

= anything having 0 in its power, it is equal to 1

Now,

x^(0) = 1



i hope this help you


-by ABHAY

Answer 2

Hey friend,
Here is the answer you were looking for:
$1) \frac{2 - \sqrt{5} }{2 + 3 \sqrt{5} } = a \sqrt{5} + b \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{2 - \sqrt{5} }{2 + 3 \sqrt{5} } \times \frac{2 - 3 \sqrt{5} }{2 - 3 \sqrt{5} } \\ \\ = \frac{2 \times 2 - 2 \times 3 \sqrt{5} - \sqrt{5} \times 2 + \sqrt{5} \times 3 \sqrt{5} }{ {(2)}^{2} - {(3 \sqrt{5} )}^{2} } \\ \\ = \frac{4 - 6 \sqrt{5} - 2 \sqrt{5} + 15}{4 - 15} \\ \\ = \frac{19 - 8 \sqrt{5} }{ - 11} \\ \\ = \frac{ - 19 + 8 \sqrt{5} }{11} = a \sqrt{5} + b \\ \\ a = \frac{8}{11} \\ \\ b = \frac{ - 19}{11} \\ \\ 2) \: \: ( \frac{ {x}^{a} }{ {x}^{b} } )^{a + b} \times ( \frac{ {x}^{b} }{ {x}^{c} } )^{b + c} \times ( \frac{ {x}^{c} }{ {x}^{a} }) ^{c + a} \\ \\ = ( {x}^{a - b} )^{a + b} \times ( {x}^{b - c})^{b + c} \times ( {x}^{c - a} )^{c + a} \\ \\ = {x}^{ {a}^{2} - {b}^{2} } \times {x}^{ {b}^{2} - {c}^{2} } \times {x}^{ {c}^{2} - {a}^{2} } \\ \\ = {x}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } \\ \\ after \: cancelling \: we \: get \\ \\ = {x}^{0} \\ \\ = 1$

Hope this helps!!!

@Mahak24

Thanks...
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