Question

Resolva a equaçao literal de incognita x :(a²+b²)x=ab (x² + 1), com a ≠ 0 e b≠0

Answer 1

$(a^2+b^2)x=ab(x^2 + 1)$
$(a^2+b^2)x=abx^2 + ab$
$abx^2-(a^2+b^2)x + ab=0$
$(ax-b)(bx-a)=0$

$ax-b=0$
$ax=b$
$\boxed{x=\frac{b}{a}}$

$ab-a=0$
$bx=a$
$\boxed{x=\frac{a}{b}}$

Answer 2

Resposta:

Explicação passo-a-passo:

(a² + b²)x = ab(x² + 1)

(a² + b²)x = abx² + ab

abx² - (a² + b²)x + ab = 0

Δ = [-(a² + b²)]² - 4 . ab . ab

Δ = a⁴ + 2a²b² + b⁴ - 4a²b²

Δ = a⁴ - 2a²b² + b⁴

Δ = (a² - b²)²

x = {-[-(a² + b²)] ± √(a² - b²)²} / (2ab)

x = [a² + b² ± (a² - b²)] / (2ab)

x' = (a² + b² + a² - b²) / (2ab)

x' = (2a²) / (2ab)

x' = a / b

x" = [a² + b² - (a² - b²)] / (2ab)

x" = (a² + b² - a² + b²) / (2ab)

x" = (2b²) / (2ab)

x" = b / a