Question

решите уравнение 1/(x-3)^2-3/x-3-4=0

Answer 1

$\frac{1}{(x-3)^2} - \frac{3}{x-3} -4=0$

$\frac{1}{(x-3)^2} -3* \frac{1}{x-3} -4=0$

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

$t^2-3t-4=0$

$D=(-3)^2-4*(-4)=25$

$t_1= \frac{3+5}{2} =4$

$t_2= \frac{3-5}{2} =-1$

$\frac{1}{x-3}=4$

$1=4(x-3)$

$4x-12=1$

$4x=13$

$x=3 \frac{1}{4}$


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

$1=-(x-3)$

$1=-x+3$

$x=2$