Question

(16. If x is a positive real number and exponents are rational numbers,
simplify

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Answer 1

Answer:

Solution

To find the value of

[\frac{x^{b} }{x^{c} } ]^{b+c -a} .[\frac{x^{c} }{x^{a} } ]^{c+a-b} .[ \frac{x^{a} }{x^{b} } ]^{a+b-c}

Here , we are applying property of exponent

\frac{x^{a} }{x^{b} } = x^{a-b}

Therefore on applying this we have

[{x^{b-c} }]^{b+c -a} .[{x^{c-a} } ]^{c+a-b} .[ {x^{a-b} ]^{a+b-c}

This can be simplified as

= [ x^{(b+c)(b-c)} .x^{-a(b-c)} ] .[ x^{(c+a)(c-a)} .x^{-b(c-a)} ] .[ x^{(a+b)(a-b)} .x^{-c(a-b)} ]

Here we are applying a^2 - b^2 = {a-b}{a+b}

We have

= [ x^{(b^{2} -c^{2} )} .x^{-ab+ac} ] .[ x^{(c^{2} -a^{2} )} .x^{-bc+ab} ] .[ x^{(a^{2} -b^{2} )} .x^{-ac+bc} ]

Again using properties of exponent we have

= [ x^{(b^{2} -c^{2} +c^{2} -a^{2} + a^{2} -b^{2} )} ] .[ x^{(-bc+ab-ab-ca+bc+ca)} ]

= [ x^{0} ] .[ x^{0} ]

= 1 . 1

= 1

Hence , finally we get 1

Answer 2

Answer:

we will apply properties of triangle then we will get ans 1