Question

Помогите пожалуйста (

Решите уравнение

а) x2-2x/x-1 - 2x-1/1-x = 3
б) x2-2x+1/x-3 + x+1/3-x = 4
в) 2/x-4 + 4/x2-4x = 0,625​

Answer 1

Ответ:

$a)\ \ \dfrac{x^2-2x}{x-1}-\dfrac{2x-1}{1-x}=3\ \ ,\ \ \ \ ODZ:\ x\ne 1\ ,\\\\\\\dfrac{x^2-2x}{x-1}+\dfrac{2x-1}{x-1}=3\ \ ,\ \ \ \dfrac{x^2-2x+2x-1}{x-1}=3\ \ ,\ \ \dfrac{x^2-1}{x-1}=3\ \ ,\\\\\\x^2-1=3(x-1)\ ,\ \ \ x^2-1=3x-3\ \ ,\ \ \ x^2-3x+2=0\ \ \ ,\\\\x_1=1\notin ODZ\ \ ,\ x_2=2\ \ (teorema\ Vieta)\\\\Otvet:\ \ x=2\ .$

$b)\ \ \dfrac{x^2-2x+1}{x-3}+\dfrac{x+1}{3-x}=4\ \ ,\ \ \ \ ODZ:\ x\ne 3\ ,\\\\\\\dfrac{x^2+2x+1}{x-3}-\dfrac{x+1}{x-3}=4\ \ ,\ \ \ \dfrac{x^2+2x+1-x-1}{x-3}=4\ \ ,\ \ \dfrac{x^2+x}{x-3}=4\ ,\\\\\\x^2+x=4(x-3)\ \ ,\ \ \ x^2-3x+12=0\ \ ,\ \ D=-39<0\ \ \to \ \ x\in \varnothing$

$c)\ \ \dfrac{2}{x-4}+\dfrac{4}{x^2-4x}=0,625\ \ ,\ \ \ \ ODZ:\ x\ne 0\ ,\ x\ne 4\ ,\\\\\\\dfrac{2}{x-4}+\dfrac{4}{x\, (x-4)}=\dfrac{5}{8}\ \ ,\ \ \ \dfrac{2x+4}{x\, (x-4)}=\dfrac{5}{8}\ \ ,\ \ \ 16x+32=5x(x-4)\ \ ,\\\\\\16x+32=5x^2-20x\ \ ,\ \ \ 5x^2-36x-32=0\ \ ,\ \ D/4=484=22^2\ ,\\\\x_1=\dfrac{6-22}{5}=-3,2\ \ \ ,\ \ \ x_2=\dfrac{6+22}{5}=5,6$