Equazione di primo grado

ricorda 2^-1 = (1/2)^1; 2^-2 = (1/2) ^2 = 1/4;

L'esponente negativo fa invertire la base e l'esponente diventa positivo.

[x + 2 + (1/2)] /[2 + (1/2)]  - [x - 2 - (1/2)] / [2 - (1/2)] = [(1/2 - 1/4) /(1/2 + 1/4)]^-1;

[x + (5/2)] / [5/2] - [x - (5/2)] / [3/2] = [(1/4) / (3/4)]^-1;

 2 * [x + 5/2] /5 - 2 * [x - (5/2)] / 3 = [(3/4) : (1/4)];

[2x + 5] / 5 - [2x - 5] / 3 = (3/4) * (4/1)

[2x + 5] / 5 - [2x - 5] / 3 = 3/1;

mcm = 3 * 5 = 15;

6x + 15 - [10x - 25] = 3 * 15;

6x + 15 - 10x + 25  = 45;

- 4x = 45 - 15 - 25;

- 4x = 5;

x = - 5/4.

@aiuto-2   ciao.

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$\small \dfrac{x+2+2^{-1}}{2+2^{-1}}-\dfrac{x-2-2^{-1}}{2-2^{-1}}= \left(\dfrac{2^{-1}-2^{-2}}{2^{-1}+2^{-2}}\right)^{-1}$

$\small \dfrac{x+2+\dfrac{1}{2}}{2+\dfrac{1}{2}}-\dfrac{x-2-\dfrac{1}{2}}{2-\dfrac{1}{2}}= \dfrac{2^{-1}+2^{-2}}{2^{-1}-2^{-2}}$

$\small \dfrac{x+\dfrac{4+1}{2}}{\dfrac{4+1}{2}}-\dfrac{x-\dfrac{4}{2}-\dfrac{1}{2}}{\dfrac{4-1}{2}}= \dfrac{\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}{\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2}$

$\small \dfrac{x+\dfrac{5}{2}}{\dfrac{5}{2}}-\dfrac{x-\dfrac{5}{2}}{\dfrac{3}{2}}= \dfrac{\dfrac{1}{2}+\dfrac{1}{4}}{\dfrac{1}{2}-\dfrac{1}{4}}$

$\small \dfrac{2}{5}\left(x+\dfrac{5}{2}\right)-\dfrac{2}{3}\left(x-\dfrac{5}{2}\right)= \dfrac{\dfrac{2+1}{4}}{\dfrac{2-1}{4}}$

$\small \dfrac{2}{5}x+\dfrac{\cancel{10}^1}{\cancel{10}_1}-\dfrac{2}{3}x+\dfrac{\cancel{10}^5}{\cancel6_3}= \dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}$

$\small \dfrac{2}{5}x+1-\dfrac{2}{3}x+\dfrac{5}{3}= \dfrac{3}{\cancel4_1}·\dfrac{\cancel4^1}{1}$

$\small \dfrac{2}{5}x+1-\dfrac{2}{3}x+\dfrac{5}{3}= 3$

$\small \dfrac{2}{5}x-\dfrac{2}{3}x= 3-1-\dfrac{5}{3}$

$\small \dfrac{2}{5}x-\dfrac{2}{3}x= 2-\dfrac{5}{3}$

$\small \dfrac{2}{5}x-\dfrac{2}{3}x= \dfrac{6-5}{3}$

$\small \dfrac{2}{5}x-\dfrac{2}{3}x= \dfrac{1}{3} \quad\color{blue}(mcm=15)$

$\small 6x-10x= 5 $

$\small -4x= 5 $

$\small \dfrac{\cancel{-4}x}{\cancel{-4}}= \dfrac{5}{-4} $

$\small x= -\dfrac{5}{4} $