aiuto equazione fratta con denominatori diversi

3·x/(x + 2) + 2·x/(x - 7) = (5·x + 6)/(x + 2)

C.E. (x + 2)·(x - 7) ≠ 0---> x ≠ 7 ∧ x ≠ -2

quindi:

3·x·(x - 7) + 2·x·(x + 2) = (5·x + 6)·(x - 7)

(3·x^2 - 21·x) + (2·x^2 + 4·x) = 5·x^2 - 29·x - 42

5·x^2 - 17·x - (5·x^2 - 29·x - 42) = 0

12·x + 42 = 0--->x = - 7/2

2x /(x-7) = (2x+6)/(x+2)

2x^2+4x = 2x^2+6x-14x-42

2x^2 si semplifica

42 = 6x-14x-4x

x = -42/12 = -7/2 

...non così difficile, al fine 😉

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$\dfrac{3x}{x+2}+\dfrac{2x}{x-7}=\dfrac{5x+6}{x+2}$ $(mcm=(x+2)(x-7)$

$3x(x-7)+2x(x+2)=(5x+6)(x-7)$

$3x^2-21x+2x^2+4x = 5x^2-35x+6x-42$

$5x^2-17x=5x^2-29x-42$

$\cancel{5x^2}-\cancel{5x^2}-17x+29x = -42$

$12x = -42$

$\dfrac{\cancel{12}x}{\cancel{12}} = \dfrac{-\cancel{42}^7}{\cancel{12}_2}$

$x= -\dfrac{7}{2}$