A awesome mathematics problem | Olympiad Question | can you solve this problem | x=?

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@tgx3529

But I tried to use substitution sqrtx +1= k
There Is the k^2=2
k= +- sqrt2
But k<0 Is not possible
For k=-sqrt2 Is x= 3+2sqrt2, Its not possible
Only 3-2sqrt2 Is possible

tgx
Автор

{2x^2/1x^2+2x^2/x^2}= 4x^4/1x^4 =4x^1 2^2x1^1 2^1^2x 1^1^2x 1^2x (x ➖ 2x+1). {2+2 ➖}/{x+x ➖ }{1+1 ➖}= {4/x^2+2 }= 6/x^2=3 (x ➖ 3x+3).

RealQinnMalloryu
Автор

_(√x + 1)²/(x - 1) + (√x + 1)/x = 2/√x + 1_
⇒ _(√x + 1)²/[(√x - 1)(√x + 1)] + (√x + 1)/x = 2/√x + 1_
⇒ _(√x + 1)/(√x - 1) + (√x + 1)/x - 2/√x - 1 = 0_
⇒ _((√x - 1)+ 2)/(√x - 1) + 1/√x + 1/x - 2/√x - 1 = 0_
⇒ _1 + 2/(√x - 1) + 1/√x + 1/x - 2/√x - 1 = 0_
⇒ _2/(√x - 1) - 1/√x + 1/x = 0_
Multiply through by _x(√x - 1)_ :
⇒ _2x - √x(√x - 1) + (√x - 1) = 0_
⇒ _x + 2√x - 1 = 0_
⇒ _(√x + 1)² - 2 = 0_
⇒ _√x + 1 = ±√2_
⇒ *_x = (±√2 - 1)² = 3 ∓ 2√2_*

guyhoghton
Автор

(√x + 1)²/(x - 1) + (√x + 1)/x = 2/√x + 1

find x ∈ ℝ
x > 0 ∧ x ≠ 1

x - 1 = (√x + 1)(√x - 1)

(√x + 1)/(√x - 1) + (√x + 1)/x = 2/√x + 1

(√x + 1)(1/(√x - 1) + 1/x) = (√x + 2)/√x

(√x + 1)(x + √x - 1)/[x(√x - 1)] = (√x + 2)/√x

(x + √x)(x + √x - 1) = x(√x - 1)(√x + 2)

(x + √x)(x + √x - 1) = x(x + √x - 2)

(x + √x)² - (x + √x) = x² + x√x - 2x
x² + x + 2x√x - x - √x = x² + x√x - 2x
x² + x√x = x² + x√x - 2x
x = 0 [ not valid ]

SidneiMV