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Question Number 142971 by 0731619 last updated on 08/Jun/21

Commented by wassel last updated on 11/Jun/21

((√x)−(1/(√x)))^2 =x+(1/x)−2=(1/2)−2=−(3/2) =(3/2)i^2   (√x)−(1/(√x))=±i(√((3/2)  ))

(x1x)2=x+1x2=122=32=32i2x1x=±i32

Answered by Rasheed.Sindhi last updated on 08/Jun/21

x+(1/x)=(1/2) ; (√x)−(1/( (√x)))=?  ((√x))^2 +(1/(((√x))^2 ))=(1/2)  ((√x))^2 +(1/(((√x))^2 ))−2=(1/2)−2=((1−4)/2)  ((√x)−(1/( (√x))))^2 =−(3/2)=−(6/4)=(±((i(√6))/2))^2   (√x)−(1/( (√x)))=±((i(√6))/2)

x+1x=12;x1x=?(x)2+1(x)2=12(x)2+1(x)22=122=142(x1x)2=32=64=(±i62)2x1x=±i62

Commented by 0731619 last updated on 08/Jun/21

tanks

tanks

Answered by Rasheed.Sindhi last updated on 08/Jun/21

((√x)−(1/( (√x))))^2 =x+(1/x)−2=(1/2)−2=−(3/2)                           =((−6)/4)=(±((i(√6))/2))^2   (√x)−(1/( (√x)))=±((i(√6))/2)

(x1x)2=x+1x2=122=32=64=(±i62)2x1x=±i62

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