Math, asked by Aishi30, 14 hours ago

answer this question​

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Answered by Anonymous
2

Answer:

let  \:  \:  \:  \frac{a}{b}  =  \frac{c}{d}  =  \frac{e}{f}  = k

a = bk \:  \:  \:  \:  \: c = dk \:  \:  \:  \:  \: e = fk

 \frac{2a {}^{4}b {}^{2}  + 3a {}^{2}c {}^{2}   - 5e {}^{4} f }{2b {}^{6}  + 3b {}^{2} d {}^{2} - 5f {}^{5}  }  =(\frac{a}{b} ) {}^{n}  \\  \frac{2b {}^{4} k {}^{4 }b {}^{2} + 3b {}^{2} k {}^{2}  d {}^{2}  k {}^{2} - 5f {}^{4}  k {}^{4}f  }{2b {}^{6}  + 3b {}^{2} d {}^{2}  - 5f {}^{5} }  =  (\frac{a}{b} ) {}^{n}  \\  \frac{2b {}^{6}k {}^{4}  + 3b {}^{2}d {}^{2}  k {}^{4}  - 5f {}^{5}k {}^{4}   }{2b {}^{6 }  + 3b {}^{2} d {}^{2}  - 5f {}^{5} }  =  (\frac{a}{b} ) {}^{n}  \\ k {}^{4}  = ( \frac{a}{b} ) {}^{n}

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