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We can define the Domain of x as followed:x + 1 ≥ 0 and x^2 - 1 ≥ 0 => x ≥ -1 and (x ≥ 1 or x ≤ -1 ) => x = -1 or x ≥ 18:43 x = 0, (1 - √5)/2 can be discarded immediately.
Yes, you are right! ❤
Let y = sqrt(x+1) then y>0y^2 = x+1x^2 - 1 = (x+1)•[(x+1) - 2]=(y^2)•(y^2 - 2) = (y^2 - 1)^2 - 1 (y^2 - 1)^2 - 1 = y(y^2 - 1)^2 = y + 1(y +1)•[(y - 1)^2] = 1[(y - 1) + 2]•[(y - 1)^2] = 1(y - 1)^3 + 2•[(y - 1)^2] = 1
Very Nice! ❤
X=-1; 0=0
Very nice! ❤
Y=Sqrt（X+1）， Y>=0, X>=-1, |X|>=0, Y^2=X+1, X^2-1=Y, X^-Y^2-1=Y-X-1, (X-Y)(X+Y+1)=0, X+Y+1>0, X=Y, X^2=X+1, X^2-X-1=0, X=(1+Sqrt (5))/2, Or X=(1-Sqrt (5))/2, refuse
x^2-1>=0 , x+1>=0 , x^2>=1 , x>=-1(x-1)(x+1)=√x+1 , √x+1=a so(x-1)a^2=a , (x-1)a^2-a=0a[(x-1)a-1]=0 , a=0 x=-1 acceptable(x-1)a-1=0 , a=1/x-1 , √x+1=1/x-1(√x+1=1/x-1)^2 , x^3-x^2-x=0x(x^2-x-1)=0 , x=0 unacceptablex^2-x-1=0 , x=(1+√5)/2 acceptable
x² - 1 = √(x + 1) → where: (x² - 1) ≥ 0 → x ⋲ ]- ∞ ; - 1] U [1 ; + ∞[[x² - 1]² = [√(x + 1)]²x⁴ - 2x² + 1 = x + 1x⁴ - 2x² - x = 0(x⁴ - 2x²) - x = 0(x² - 1)² - 1 - x = 0(x² - 1)² - (x + 1) = 0[(x + 1).(x - 1)]² - (x + 1) = 0(x + 1)².(x - 1)² - (x + 1) = 0(x + 1).[(x + 1).(x - 1)² - 1] = 0(x + 1).[(x + 1).(x² - 2x + 1) - 1] = 0(x + 1).[x³ - 2x² + x + x² - 2x + 1 - 1] = 0(x + 1).[x³ - x² - x] = 0x.(x + 1).(x² - x - 1) = 0First case: x = 0 → rejected, because the conditionSecond case: (x + 1) = 0x + 1 = 0x = - 1Third case: (x² - x - 1) = 0x² - x - 1 = 0Δ = (- 1)² - (4 * - 1) = 1 + 4 = 5x = (1 ± √5)/2First possibility:x = (1 - √5)/2 → rejected, because the conditionSecond possibility:x = (1 + √5)/2Solution = { - 1 ; (1 + √5)/2 }
Минус один, решил за 30 секунд подбором
Очень хорошо! ❤
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We can define the Domain of x as followed:
x + 1 ≥ 0 and x^2 - 1 ≥ 0 => x ≥ -1 and (x ≥ 1 or x ≤ -1 ) => x = -1 or x ≥ 1
8:43 x = 0, (1 - √5)/2 can be discarded immediately.
Yes, you are right! ❤
Let y = sqrt(x+1) then y>0
y^2 = x+1
x^2 - 1 = (x+1)•[(x+1) - 2]
=(y^2)•(y^2 - 2) = (y^2 - 1)^2 - 1
(y^2 - 1)^2 - 1 = y
(y^2 - 1)^2 = y + 1
(y +1)•[(y - 1)^2] = 1
[(y - 1) + 2]•[(y - 1)^2] = 1
(y - 1)^3 + 2•[(y - 1)^2] = 1
Very Nice! ❤
X=-1; 0=0
Very nice! ❤
Y=Sqrt（X+1）， Y>=0, X>=-1, |X|>=0, Y^2=X+1, X^2-1=Y, X^-Y^2-1=Y-X-1, (X-Y)(X+Y+1)=0, X+Y+1>0, X=Y, X^2=X+1, X^2-X-1=0, X=(1+Sqrt (5))/2, Or X=(1-Sqrt (5))/2, refuse
Very nice! ❤
x^2-1>=0 , x+1>=0 , x^2>=1 , x>=-1
(x-1)(x+1)=√x+1 , √x+1=a so
(x-1)a^2=a , (x-1)a^2-a=0
a[(x-1)a-1]=0 , a=0 x=-1 acceptable
(x-1)a-1=0 , a=1/x-1 , √x+1=1/x-1
(√x+1=1/x-1)^2 , x^3-x^2-x=0
x(x^2-x-1)=0 , x=0 unacceptable
x^2-x-1=0 , x=(1+√5)/2 acceptable
Very nice! ❤
x² - 1 = √(x + 1) → where: (x² - 1) ≥ 0 → x ⋲ ]- ∞ ; - 1] U [1 ; + ∞[
[x² - 1]² = [√(x + 1)]²
x⁴ - 2x² + 1 = x + 1
x⁴ - 2x² - x = 0
(x⁴ - 2x²) - x = 0
(x² - 1)² - 1 - x = 0
(x² - 1)² - (x + 1) = 0
[(x + 1).(x - 1)]² - (x + 1) = 0
(x + 1)².(x - 1)² - (x + 1) = 0
(x + 1).[(x + 1).(x - 1)² - 1] = 0
(x + 1).[(x + 1).(x² - 2x + 1) - 1] = 0
(x + 1).[x³ - 2x² + x + x² - 2x + 1 - 1] = 0
(x + 1).[x³ - x² - x] = 0
x.(x + 1).(x² - x - 1) = 0
First case: x = 0 → rejected, because the condition
Second case: (x + 1) = 0
x + 1 = 0
x = - 1
Third case: (x² - x - 1) = 0
x² - x - 1 = 0
Δ = (- 1)² - (4 * - 1) = 1 + 4 = 5
x = (1 ± √5)/2
First possibility:
x = (1 - √5)/2 → rejected, because the condition
Second possibility:
x = (1 + √5)/2
Solution = { - 1 ; (1 + √5)/2 }
Very nice! ❤
Минус один, решил за 30 секунд подбором
Очень хорошо! ❤
X=-1; 0=0
Very nice! ❤