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$\dfrac{(x+1)(x-1)}{3}-\dfrac{1}{3}(x-2)^2+\dfrac{2x-1}{4} = \dfrac{2}{3}-\dfrac{x+1}{4}-\dfrac{23}{12}$
mcm dei denominatori = 12 quindi moltiplica tutto per 12 così li elimini:
$4(x+1)(x-1)-4(x-2)^2+3(2x-1) = 8-3(x+1)-23$
$4(x^2-1)-4(x^2-4x+4)+6x-3 = 8-3x-3-23$
$4x^2-4-4x^2+16x-16+6x-3 = -3x-18$
$22x-23 = -3x-18$
$22x+3x = -18+23$
$25x = 5$
dividi ambo le parti per 5:
$5x = 1$
$x= \frac{1}{5}$
(x + 1)·(x - 1)/3 - 1/3·(x - 2)^2 + (2·x + 1)/4 = 2/3 - (x + 1)/4 - 23/12
(x^2 - 1)/3 - 1/3·(x^2 - 4·x + 4) + (2·x + 1)/4 = 2/3 - (x + 1)/4 - 23/12
4·(x^2 - 1) - 4·(x^2 - 4·x + 4) + 3·(2·x + 1) = 8 - 3·(x + 1) - 23
(4·x^2 - 4) - (4·x^2 - 16·x + 16) + (6·x + 3) = 8 - (3·x + 3) - 23
22·x - 17 = - 3·x - 18
25·x = -1
x = - 1/25