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ICSE Class 9 Answered

Question no. 7c
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Asked by jha.pooja1702 | 11 Feb, 2018, 11:11: PM
answered-by-expert Expert Answer
begin mathsize 16px style open parentheses straight x plus straight y close parentheses to the power of negative 1 end exponent open parentheses straight x to the power of negative 1 end exponent plus straight y to the power of negative 1 end exponent close parentheses equals straight x to the power of straight p straight y to the power of straight q
fraction numerator 1 over denominator straight x plus straight y end fraction cross times open parentheses 1 over straight x plus 1 over straight y close parentheses equals straight x to the power of straight p straight y to the power of straight q
fraction numerator 1 over denominator straight x plus straight y end fraction cross times open parentheses fraction numerator straight x plus straight y over denominator xy end fraction close parentheses equals straight x to the power of straight p straight y to the power of straight q
1 over xy equals straight x to the power of straight p straight y to the power of straight q
straight x to the power of negative 1 end exponent straight y to the power of negative 1 end exponent equals straight x to the power of straight p straight y to the power of straight q
straight p equals negative 1 space and space straight q equals negative 1
Consider comma
straight p plus straight q plus 2 equals negative 1 minus 1 plus 2 equals 0
Hence space proved. end style
Answered by Sneha shidid | 12 Feb, 2018, 09:48: AM
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