Question

determine o valor de "x" resolvendo a equacao fracionaria.​

Answer 1

$\frac{1}{x-1}-\frac{2}{x-2}+\frac{1}{x+3}=0$

$\frac{(x-2)(x+3)-2(x-1)(x+3)+(x-1)(x-2)}{(x-1)(x-2)(x+3)}=0$

${x}^{2}+x-6-2({x}^{2}+2x-3)+{x}^{2}-3x+2=0$

${x}^{2}+x-6-2{x}^{2}-4x+6+{x}^{2}-3x+2=0$

$x-4x-3x+2=0$

$-6x=-2\times(-1)$

$6x=2$

$x=\frac{2}{6}$

$\boxed{x=\frac{1}{3}}$

Answer 2

multiplicando toda a expressão por (x-1)(x-2)(x+3), teremos

$(x - 2)(x + 3) - 2(x - 1)(x + 3) + (x - 1)(x - 2) = 0$

${x}^{2} + x - 6 - 2 {x}^{2} - 4x + 6 + {x}^{2} - 3x + 2 = 0$

$- 6x + 2 = 0$

$- 6x = - 2$

$x = \frac{ - 2}{ - 6}$

$x = \frac{1}{3}$

x=1/3