[Risolto] n71

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71)

$\dfrac{3}{2}x + \dfrac{5-x}{12} = \dfrac{(2x-1)5}{6}-\dfrac{x(x+1)}{4}$

$\dfrac{3}{2}x + \dfrac{5-x}{12} = \dfrac{10x-5}{6}-\dfrac{x^2+x}{4}$

mcm= 12:

$18x +5-x = 2(10x-5) -3(x^2+x)$

$18x +5 -x = 20x -10 -3x^2 -3x$

$17x +5 = 17x -10 -3x^2$

$3x^2 \cancel{+17x} \cancel{-17x}  = -10 -5$

$3x^2 = -15$

$x^2 = -5$

$\sqrt{x^2} = \sqrt5 ×\sqrt{-1}$

per cui nei complessi $ℂ:$

$x= \pm\approx{2,236068i}.$