A nice Math Olympiad Algebra Equation. Can you dare solve this ? Find the Value of x ? 

Your videos help me to think differently about equations. Im learning a lot. Thank you so much.

Very good solution.many many thanks sir.❤

Good

Very good work

Good solution, but in this case it seems easier just to subtract the terms, cross multiply, expand and simplify. We have _(⁽ˣ⁺¹⁾/ₓ)² - (⁽ˣ⁺¹⁾/₍ₓ₋₁₎)² = 1_ ⇒ _(x + 1)²[⅟ₓ² - ⅟₍ₓ₋₁₎²] = 1_ ⇒ _(x + 1)²[(x - 1)² - x²]/[x²(x - 1)²] = 1_ ⇒ _(x + 1)²(1 - 2x) = x²(x - 1)²_ ⇒ _-2x³ - 3x² + 1 = x⁴ - 2x³ + x²_ ⇒ _(x²)² + 4x² - 1 = 0_ ⇒ _x² = -2 ± √5_ ⇒ *_x = ±√(-2 ±√5)_*

Use ((x+1)/x)² = =((x+1)/(x-1))²+1 =2(x²+1)/(x-1)² Simplify it and get (x² -1)² =2x²(x²+1) => X⁴+4x²-1=0 => x²=-2±5½. So x=±(-2+5½)½, ±(2+5½)½ i.

The equation to solve is ((x + 1)/x)² − ((x + 1)/(x − 1))² = 1 Start by multiplying both sides by x²(x − 1)² to get rid of the fractions and we have (x + 1)²(x − 1)² − (x + 1)²x² = x²(x − 1)² (x + 1)²(x − 1)² = (x + 1)²x² + x²(x − 1)² (x² − 1)² = x²((x + 1)² + (x − 1)²) (x² − 1)² = x²(2x² + 2) x⁴ − 2x² + 1 = 2x⁴ + 2x² x⁴ + 4x² − 1 = 0 x² = −2 + √5 ⋁ x² = −2 − √5 x = √(−2 + √5) ⋁ x = −√(−2 + √5) ⋁ x = i√(2 + √5) ⋁ x = −i√(2 + √5)

Easy way to solve this equation Equation has solution for x not equal to 0 or 1 (X+1)^2/((1/X^2-1/(X-1)^2) =1 Or, ( (X+1)^2)((X-1)^2-x^2))/((X^2)(X-1)^2) =1 Or, ((X+1)^2)(X-1+X)(X-1-X) =(X^2)(X-1)^2 Or, (X^2+2X+1)(2X-1)(-1) =(X^2)(X^2-2X+1) Or, (X^2+2X+1)(1-2X) =X^4-2X^3+X^2 Or, X^2+2X+1-2X^3-4X^2-2X = X^4-2X^3+X^2 Simplifying we get X^4+4X^2-1 =0 Put X^2=t Then X^4=t^2 Equation t^2+4t-1=0 t=-2+√5 or t = -2-√5 Case 1 X^2=-2+√5 X1= √(-2+√5) or X2 = -√(-2+√5) Case 2 X^2= -2-√5 =(-1)(2+√5) X3=i√(2+√5) or X4= -i√(2+√5)

If I put for x=-0,5, in the equation, then the solution of the equation is 1,11'. For this I try with the number -0,485 and put it for x. Then the solution of the equation is: [(-0,485+1)^2/-0,485^2] - [(-0,485+1)^2/(-0,485-1)^2 =1,006 x=-0,485

A nice Math Olympiad Algebra Equation: [(x + 1)/x]² - [(x + 1)/(x - 1)² = 1; x = ? x ≠ 0, x ≠ 1; (x + 1)²[(x - 1)² - x²]/[(x²)(x - 1)²] = 1 [(x + 1)²(1 - 2x)]/[(x²)(x - 1)²] = (1 - 3x² - 2x³)/(x⁴ - 2x³ + x²) = 1 1 - 3x² - 2x³ = x⁴ - 2x³ + x², x⁴ + 4x² - 1 = 0, (x² + 2)² = 5 = (√5)² x² = - 2 + √5 or x² = - 2 - √5 = - (2 + √5); Imaginary value roots x² = - 2 + √5 = 0.236; x = ± √(- 2 + √5) = ± √0.236 = ± 0.486 x² = - (2 + √5) = - 4.236; x = ± i√(2 + √5) = ± i√4.236 = ± 2.058i The calculation was achieved on a smartphone with a standard calculator app Answer check: [(x + 1)/x]² - [(x + 1)/(x - 1)]² = (1 - 3x² - 2x³)/(x⁴ - 2x³ + x²) x² = - 2 ± √5: x⁴ + 4x² - 1 = 0, x⁴ = 1 - 4x² (1 - 3x² - 2x³)/(x⁴ - 2x³ + x²) = (1 - 3x² - 2x³)/(1 - 4x² - 2x³ + x²) = (1 - 3x² - 2x³)/(1 - 3x² - 2x³) = 1; Confirmed Final answer: x = √(- 2 + √5) = 0.486, x = - √(- 2 + √5) = - 0.486 x = i√(2 + √5) = 2.058i or x = - i√(2 + √5) = - 2.058i