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Question Number 174084 by AgniMath last updated on 24/Jul/22

Commented by infinityaction last updated on 24/Jul/22

(((√(3+x))+(√(3−x))+(√(3+x))−(√(3−x)))/( (√(3+x))+(√(3−x))−(√(3+x))+(√(3−x)))) = ((2+1)/(2−1)) ((√(3+x))/( (√(3−x)))) = 3 ((3+x)/(3−x)) = 9 (6/(2x)) = ((10)/(8 )) ⇒ x = ((24)/(10)) x = ((12)/5)

3+x+3−x+3+x−3−x3+x+3−x−3+x+3−x=2+12−13+x3−x=33+x3−x=962x=108⇒x=2410x=125

Answered by puissant last updated on 24/Jul/22

x ≥ −3 ; x ≤ 3 ⇒ −3≤x≤3. ((((√(3+x))+(√(3−x)))^2 )/(3+x−3+x )) = 2 ⇔ ((3+x+2(√(9−x^2 ))+3−x)/(2x)) = 2 ⇔ (√(9−x^2 )) = 2x−3 ⇔ 9−x^2 =4x^2 −12x+9 ⇔ 5x^2 −12x=0 ⇔ x=((12)/5)...

x⩾−3;x⩽3⇒−3⩽x⩽3.(3+x+3−x)23+x−3+x=2⇔3+x+29−x2+3−x2x=2⇔9−x2=2x−3⇔9−x2=4x2−12x+9⇔5x2−12x=0⇔x=125...

Commented by som(math1967) last updated on 24/Jul/22

if x=0 then (((√3)+(√3))/( (√3)−(√3))) ≠2

ifx=0then3+33−3≠2

Answered by Rasheed.Sindhi last updated on 24/Jul/22

(((√(3+x)) +(√(3−x)) )/( (√(3+x)) −(√(3−x)) ))=2 (√(3+x)) +(√(3−x)) =2(√(3+x)) −2(√(3−x)) (√(3+x)) =3(√(3−x)) 3+x=9(3−x) 3+x=27−9x x=((24)/(10))=((12)/5)

3+x+3−x3+x−3−x=23+x+3−x=23+x−23−x3+x=33−x3+x=9(3−x)3+x=27−9xx=2410=125

Commented by AgniMath last updated on 25/Jul/22

Thanks for the easiest solution

Thanksfortheeasiestsolution

Commented by Rasheed.Sindhi last updated on 26/Jul/22

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