Urgente

(3·x - 5)·(x + 1)/18 + (x + 2)·(x + 3)/6 = (2·x + 3)^2/12

(2·(3·x - 5))·(x + 1) + (6·(x + 2))·(x + 3) = 3·(2·x + 3)^2

(6·x - 10)·(x + 1) + (6·x + 12)·(x + 3) = 3·(4·x^2 + 12·x + 9)

(6·x^2 - 4·x - 10) + (6·x^2 + 30·x + 36) = 12·x^2 + 36·x + 27

26·x + 26 = 36·x + 27

10·x = -1----> x = - 1/10

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(x - 1)^2/4 - (x + 1)^2/3 = (1 + x)·(1 - x)/12 - (x - 2)/6

3·(x - 1)^2 - 4·(x + 1)^2 = (1 + x)·(1 - x) - 2·(x - 2)

3·(x^2 - 2·x + 1) - 4·(x^2 + 2·x + 1) = (1 - x^2) - (2·x - 4)

(3·x^2 - 6·x + 3) - (4·x^2 + 8·x + 4) = (1 - x^2) - (2·x - 4)

- x^2 - 14·x - 1 = - x^2 - 2·x + 5

- 14·x - 1 = - 2·x + 5

- 12·x = 6----> x = - 1/2

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$\small \dfrac{(3x-5)(x+1)}{18}+\dfrac{(x+2)(x+3)}{6} = \dfrac{(2x+3)^2}{12}$

$\small \dfrac{3x^2+3x-5x-5}{18}+\dfrac{x^2+3x+2x+6}{6} = \dfrac{4x^2+12x+9}{12}$

$\small \dfrac{3x^2-2x-5}{18}+\dfrac{x^2+5x+6}{6} = \dfrac{4x^2+12x+9}{12}$ $\quad \small [mcm=36]$

$\small 2(3x^2-2x-5)+6(x^2+5x+6) = 3(4x^2+12x+9)$

$\small 6x^2-4x-10+6x^2+30x+36 = 12x^2+36x+27$

$\small 12x^2+26x+26 = 12x^2+36x+27$

$\small \cancel{12x^2}+26x-\cancel{12x^2}-36x = 27-26$

$\small -10x = 1$

$\small 10x = -1$

$\small \dfrac{\cancel{10}x}{\cancel{10}} = \dfrac{-1}{10}$

$\small x = -\dfrac{1}{10}$