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Question Number 202638 by depressiveshrek last updated on 31/Dec/23

(√(x−(1/x)))−(√(1−(1/x)))=1−(1/x)

x1x11x=11x

Commented by deleteduser1 last updated on 31/Dec/23

x=1.Suppose x≠1,let a=(√(x−(1/x)));b=(√(1−(1/x)))  a−b=1−(1/x);(a+b)^2 =1+x  a^2 +b^2 =x+1−(2/x)⇒2ab=(2/x)  (a+b)^2 −(a−b)^2 =4ab⇒x−(1/x^2 )+(2/x)=(4/x)  ⇒x^3 −2x−1=0⇒x=((1+(√5))/2)

x=1.Supposex1,leta=x1x;b=11xab=11x;(a+b)2=1+xa2+b2=x+12x2ab=2x(a+b)2(ab)2=4abx1x2+2x=4xx32x1=0x=1+52

Commented by Rasheed.Sindhi last updated on 31/Dec/23

Sir mr W “can see” geometry in the question.

SirmrWcanseegeometryinthequestion.

Commented by deleteduser1 last updated on 31/Dec/23

Commented by deleteduser1 last updated on 31/Dec/23

OR a^2 −b^2 =(a−b)(a+b)=x−1  ⇒a−b=((x−1)/( (√(1+x))))=((x−1)/x)⇒x=1 or x=(√(x+1))  ⇒x^2 −x−1=0⇒x=((1+(√5))/2)

ORa2b2=(ab)(a+b)=x1ab=x11+x=x1xx=1orx=x+1x2x1=0x=1+52

Answered by Rasheed.Sindhi last updated on 01/Jan/24

(√(x−(1/x)))−(√(1−(1/x)))=1−(1/x)  (√((x^2 −1)/x)) −(√((x−1)/x)) −((x−1)/x)=0  (√(((x−1)/x)(x+1))) −(√((x−1)/x)) −((√((x−1)/x)) )^2 =0  (√((x−1)/x)) ((√(x+1)) −1−(√((x−1)/x)) )=0  (√((x−1)/x)) =0 ∣ (√(x+1)) −1−(√((x−1)/x)) =0       ((x−1)/x)=0 ∣ (√(x+1)) −(√((x−1)/x)) =1       x=1✓ ∣ (x+1)+(((x−1)/x))−2(√((x^2 −1)/x)) =1  x+1+(1−(1/x))−2(√(x−(1/x))) =1  1+x−(1/x)−2(√(x−(1/x))) =0  1+((√(x−(1/x))) )^2 −2(√(x−(1/x))) =0  y^2 −2y+1=0  y=1  (√(x−(1/x))) =1  x^2 −x−1=0  x=((1±(√(1+4)) )/2)     =((1±(√5) )/2)

x1x11x=11xx21xx1xx1x=0x1x(x+1)x1x(x1x)2=0x1x(x+11x1x)=0x1x=0x+11x1x=0x1x=0x+1x1x=1x=1(x+1)+(x1x)2x21x=1x+1+(11x)2x1x=11+x1x2x1x=01+(x1x)22x1x=0y22y+1=0y=1x1x=1x2x1=0x=1±1+42=1±52

Commented by deleteduser1 last updated on 31/Dec/23

8th line: x+1+(1−(1/x))−2(√(x−(1/x)))=0

8thline:x+1+(11x)2x1x=0

Commented by Rasheed.Sindhi last updated on 31/Dec/23

Thanks sir,I′ve corrected.

Thankssir,Ivecorrected.

Commented by Rasheed.Sindhi last updated on 02/Jan/24

ReCorrected. Now the answer is  right!

ReCorrected.Nowtheanswerisright!

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