Nice problem but the solution seems to be too complicated. Put ((x+1)/(x-1))² on the RHS and multiply by x² (x - 1)²: (x + 1)² (x - 1)² = x² (x - 1)² + x² (x + 1)² x⁴ - 2x² + 1 = x² (2x² + 2) x⁴ + 4x² - 1 = 0 solve for x²: x² = -2 ± √5 now we get x = ±√( -2 ±√5)
Resolvi de maneira bem mais simples. Desenvolvi e cheguei em x^4 + 4x^2 = 1, adicionei 4 aos dois lados, fatorei e calculei as raízes. (I solved it in a much simpler way. I developed and arrived at x^4 + 4x^2 = 1, I added 4 to both sides, factored and calculated the roots.)
Your way it's ok but too hard and long. This is a short solution: (x+1/x)^2*(1-(x/x+1)^2*(x+1/x-1)^2)=1 (x+1/x)^2*(1-(x/(x-1))^2)=1 1-(x/(x-1))^2=(x/(x+1))^2 (x/(x-1))^2+(x/(x+1))^2=1 With commun dominator : x^2*(2x^2+2)/(x^2-1)^2=1 And put y=x^2 y^2+4y-1=0 y1= √5-2 et y2= -√5-2 Only y1 so X = +-√(√5-2)
Nice problem but the solution seems to be too complicated.
Put ((x+1)/(x-1))² on the RHS and multiply by x² (x - 1)²:
(x + 1)² (x - 1)² = x² (x - 1)² + x² (x + 1)²
x⁴ - 2x² + 1 = x² (2x² + 2)
x⁴ + 4x² - 1 = 0
solve for x²: x² = -2 ± √5
now we get x = ±√( -2 ±√5)
totally agreed!
nice🎉
did the same. But Super Academy seems to like to overcomplicate his solutions...
Resolvi de maneira bem mais simples. Desenvolvi e cheguei
em x^4 + 4x^2 = 1, adicionei 4 aos dois lados, fatorei e calculei as raízes. (I solved it in a much simpler way. I developed and arrived
at x^4 + 4x^2 = 1, I added 4 to both sides, factored and calculated the roots.)
I did the same, it tooks minutes five minutes.
But why make it simple when you can make it complicated😂😂
(-0,5+1/-0,5)^2 - (-0,5+1/-0,5-1)^2=
(0,25/0,25)-(0,25/2,25)=1-0,1=0,9
(-0,485+1/-0,485)^2-(-0,485+1/-0,485-1)^2=
(0,265/0,235)-(0,265/2,205)=
1,127-0,120=1,00 x=-0,485
Your way it's ok but too hard and long. This is a short solution:
(x+1/x)^2*(1-(x/x+1)^2*(x+1/x-1)^2)=1
(x+1/x)^2*(1-(x/(x-1))^2)=1
1-(x/(x-1))^2=(x/(x+1))^2
(x/(x-1))^2+(x/(x+1))^2=1
With commun dominator :
x^2*(2x^2+2)/(x^2-1)^2=1
And put y=x^2
y^2+4y-1=0
y1= √5-2 et y2= -√5-2
Only y1 so X = +-√(√5-2)
Thanks for the nice method 😎😍
[(x+1)/x]² - [(x+1)/(x-1)]² = 1
Let, [(x+1)/x]² be A and [(x+1)/(x-1)]² be B
A = [(x+1)/x]² = (x+1)²/x² = (x²+2x+1)/x²
B = [(x+1)/(x-1)]² = (x+1)²/(x-1)² = (x²+2x+1)/(x²-2x+1)
A-B = 1
=> (x²+2x+1)/x² - (x²+2x+1)/(x²-2x+1) = 1
=> [(x²+2x+1)(x²-2x+1) - (x²+2x+1)x²]/[x²(x²-2x+1)] = 1
L.H.S = N/D
N = (x²+2x+1)(x²-2x+1) - x²(x²+2x+1) = x⁴ - 2x³ + x² + 2x³ - 4x² + 2x + x² - 2x + 1 = x⁴ - 2x³ + 1
D = x²(x²+2x+1) = x⁴ + 2x³ + x²
N/D = [x⁴-2x²+1-(x⁴+2x³+x²)]/[x²(x²-2x+1)] = 1
=> x²(x²-2x+1) = x⁴-2x²+1-x⁴-2x³-x²
=> x⁴-2x³+x³+2x³+3x²-1=0
=> x⁴ + 4x² - 1 = 0
(suppose: y = x²)
(a=1, b=4, c =-1)
∆ = b²-4ac = (4)²-4(1)(-1) = 16 -(-4) = 16+4 = 20
y = (-b±√∆)/2a = (-4±√20)/2 = (-4 ± 2√5)/2 = -2±√5
[Recall → y² = x]
Case 1:
x² = -2 - √5
=> x² = -(2 + √5)
=> x = i²(2 + √5)
=> x= ±i√(2+√5))
Case 2:
x² = -2 + √5
=> x = ±√(-2 + √5)
Nice method 😎💯🥰
Il y a en efffet beaucoup plus simple !
[(x + 1)/x]² - [(x + 1)/(x - 1)]² = 1
[(x + 1)²/x²] - [(x + 1)²/(x - 1)²] = 1
[(x + 1)².(x - 1)² - x².(x + 1)²] / [x².(x - 1)²] = 1
(x + 1)².(x - 1)² - x².(x + 1)² = x².(x - 1)²
(x + 1)².(x - 1)² - x².(x + 1)² - x².(x - 1)² = 0
(x + 1)².(x - 1)² - x².[(x + 1)² + (x - 1)²] = 0
(x² + 2x + 1).(x² - 2x + 1) - x².[x² + 2x + 1 + x² - 2x + 1] = 0
(x⁴ - 2x³ + x² + 2x³ - 4x² + 2x + x² - 2x + 1) - x².(2x² + 2) = 0
(x⁴ - 2x² + 1) - 2x⁴ - 2x² = 0
x⁴ - 2x² + 1 - 2x⁴ - 2x² = 0
- x⁴ - 4x² + 1 = 0
x⁴ + 4x² - 1 = 0
Δ = 4² - (4 * - 1) = 16 + 4 = 20
x² = (- 4 ± √20)/2
x² = (- 4 ± 2√5)/2
x² = - 2 ± √5
First case: x² = - 2 + √5
x² = - 2 + √5
x = ± √(- 2 + √5)
Second case: x² = - 2 - √5
x² = - (2 + √5)
x² = i².(2 + √5)
x = ± i.√(√5 + 2)