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Question Number 32371 by .none. last updated on 24/Mar/18

−1⟨x⟨0 (√x^2 )−(√((x+(1/x))^2 −4))=−2x+(1/x) Why?

−1⟨x⟨0x2−(x+1x)2−4=−2x+1xWhy?

Commented by abdo imad last updated on 25/Mar/18

we have 0<−x<1 we use the ch.t=−x ⇒ 0<t<1 and prove that (√(t^2 )) −(√((t +(1/t))^2 −4)) =2t −(1/t) we have (√t^2 ) −(√((t+(1/t))^2 −4)) =t −(√(t^2 +2 +(1/t^2 )−4)) =t −(√((t−(1/t))^2 )) =t −∣t −(1/t)∣ but we have t −(1/t) =((t^2 −1)/t)<0 =t −((1/t) −t) =2t −(1/t) =−2x +(1/x) .

wehave0<−x<1weusethech.t=−x⇒0<t<1andprovethatt2−(t+1t)2−4=2t−1twehavet2−(t+1t)2−4=t−t2+2+1t2−4=t−(t−1t)2=t−∣t−1t∣butwehavet−1t=t2−1t<0=t−(1t−t)=2t−1t=−2x+1x.

Answered by MJS last updated on 24/Mar/18

I. (√x^2 )=−x with x<0 [i.e. (√((−2)^2 ))=(√4)=2] II. (√((x+(1/x))^2 −4))= [x≠0] =(√(x^2 +2+(1/x^2 )−4)) =(√(x^2 −2+(1/x^2 )))=(√((x−(1/x))^2 ))= =(x−(1/x)) with −1<x<0 [−1<x<0 ⇒ ∣x∣<1 ⇒ ∣(1/x)∣>1 ⇒ ⇒(x−(1/x))>0; i.e. −(1/2)−(1/(−(1/2)))= =−(1/2)+2=(3/2)] I.−II. −x−(x−(1/x))=−2x+(1/x)

I.x2=−xwithx<0[i.e.(−2)2=4=2]II.(x+1x)2−4=[x≠0]=x2+2+1x2−4=x2−2+1x2=(x−1x)2==(x−1x)with−1<x<0[−1<x<0⇒∣x∣<1⇒∣1x∣>1⇒⇒(x−1x)>0;i.e.−12−1−12==−12+2=32]I.−II.−x−(x−1x)=−2x+1x

Answered by $@ty@m last updated on 24/Mar/18

(√x^2 )=∣x∣=−x (∵−1<x<0) & (√((x+(1/x))^2 −4))=∣x−(1/x)∣=(1/x)−x

x2=∣x∣=−x(∵−1<x<0)&(x+1x)2−4=∣x−1x∣=1x−x

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