TheAnimeGirl
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- Simplify :

[tex] \large{ \bf { \frac{( {a}^{2} - \frac{1}{ {b}^{2} } ) ^{a \: } (a - \frac{1}{b} ) ^{b - a} }{( {b}^{2} - \frac{1}{ {a}^{2} } ) ^{b} \: (b + \frac{1}{a} ) \: {}^{a - b} } }}[/tex]

Help! Please provide the steps too!
Irrelevant / Random answers will be reported! Ty in advance! :) ​

Sagot :

mhanifa

Answer:

  • [tex](ab)^{a+b}[/tex]

Step-by-step explanation:

[tex](a^2-1/b^2)^a(a - 1/b)^{b-a}/({b^2-1/a^2)^b(b +1/a)^{a-b}) =[/tex]

[tex]((a^2b^2-1)/b^2)^a((ab-1)/b)^{b-a}/(((a^2b^2-1)/a^2)^b(ab+1)/a)^{a-b}) =[/tex] [tex](((ab+1)(ab-1))^a/b^{2a})((ab-1)^{b-a}/b^{b-a})/((((ab+1)(ab-1))^b/a^{2b})(ab+1)^{a-b}/a^{a-b})) =[/tex][tex](ab+1)^{a-b-a+b}(ab-1)^{a+b-a-b}b^{2a+b-a}a^{2b+a-b} =[/tex]

[tex](ab+1)^0(ab -1)^0a^{a+b}b^{a+b}=[/tex]

[tex](ab)^{a+b}[/tex]

VirαtKσhli

[tex]\underline{\underline{\red{\textsf {\textbf{ Given :- }}}}}[/tex]

  • [tex] \sf { \dfrac{\bigg( {a}^{2} - \dfrac{1}{ {b}^{2} } \bigg) ^{a \: } \bigg(a - \dfrac{1}{b} \bigg) ^{b - a} }{\bigg( {b}^{2} - \dfrac{1}{ {a}^{2} } \bigg) ^{b} \: \bigg(b + \dfrac{1}{a} \bigg) \: {}^{a - b} } }[/tex]

[tex]\underline{\underline{\red{\textsf{\textbf{To \ Find :- }}}}}[/tex]

  • The simplified form .

[tex]\underline{\underline{\red{\textsf {\textbf{Answer :- }}}}}[/tex]

The given expression to us is ,

[tex]\longrightarrow\footnotesize{ \sf \sf { \dfrac{\bigg( {a}^{2} - \dfrac{1}{ {b}^{2} } \bigg) ^{a \: } \bigg(a - \dfrac{1}{b} \bigg) ^{b - a} }{\bigg( {b}^{2} - \dfrac{1}{ {a}^{2} } \bigg) ^{b} \: \bigg(b + \dfrac{1}{a} \bigg) \: {}^{a - b} } } }\\\\\\ \longrightarrow\footnotesize{ \sf \dfrac{\bigg( \dfrac{a^2b^2-1}{b^2}\bigg)^a \bigg( \dfrac{ab-1}{b}\bigg)^{b-a} }{ \bigg( \dfrac{b^2a^2-1}{a^2} \bigg)^b \bigg( \dfrac{ ba+1}{a}\bigg)^{a-b} } } \\\\\\ \longrightarrow\footnotesize{ \sf \dfrac{ \dfrac{ ( ab +1)^a(ab-1)^b }{b^{2a} } .\dfrac{(ab-1)^{b-a}}{b^{b-a} }}{ \dfrac{ (ba+1)^b(ba-1)^b }{a^{2b}}. \dfrac{(ba+1)^{a-b}}{a^{a-b}} } } \\\\\\ \longrightarrow\footnotesize{ \sf \dfrac{ \dfrac{(ab+1)^a ( ab -1)^{a+b-1 } }{b^{2a + b - a }}}{ \dfrac{(ab+1)^{b-b + a } ( ab -1)^b }{ a^{2b + a - b }}}} \\\\[/tex]

[tex]\\ \longrightarrow\footnotesize{\sf \dfrac{ \cancel{(ab+1)^a ( ab -1)^b }( a^{a+b}) }{ b^{a + b } \cancel{( ab +1)^a ( ab -1)^b} } } \\\\[/tex]

[tex]\\ \longrightarrow\footnotesize{\sf \boxed{\red{\sf \bigg\lgroup \dfrac{a}{b} \bigg\rgroup ^{a+b } }}}[/tex]

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