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Ответ:

Решить уравнение .

[tex]\bf \dfrac{6-y}{1-y^2}-\dfrac{y+3}{y-y^2}=\dfrac{y+5}{y+y^2}[/tex]  

Раскладываем знаменатели на множители и приводим дроби к общему знаменателю .

[tex]\bf \dfrac{6-y}{(1-y)(1+y)}-\dfrac{y+3}{y(1-y)}=\dfrac{y+5}{y(1+y)}\ \ ,\ \ \ ODZ:\ y\ne 0\ ,\ y\ne \pm 1\ \ .\\\\\\\dfrac{(6-y)\, y-(y+3)(1+y)-(y+5)(1-y)}{y\, (1-y)(1+y)}=0\\\\\\\dfrac{6y-y^2-(y^2+4y+3)-(-y^2-4y+5)}{y(1-y)(1+y)}=0\\\\\\\dfrac{-y^2+6y-8}{y(1-y)(1+y)}=0\ \ \ \Rightarrow \ \ \ y^2-6y+8=0\ \ ,\ \ y_1=2\ ,\ y_2=4\ \ (Viet)\\\\\\Otvet:\ y_1=2\ ,\ y_2=4\ .[/tex]

Ответ: у₁=2   у₂=4.

Объяснение:

[tex]\displaystyle\\\frac{6-y}{1-y^2} -\frac{y+3}{y-y^2} =\frac{y+5}{y+y^2} \\\\\\\frac{6-y}{(1-y)(1+y)} -\frac{y+3}{y(1-y)}-\frac{y+5}{y(1+y)}=0\\\\\\ \frac{(6-y)y-(y+3)(1+y)-(y+5)(1-y)}{y(1-y)(1+y)}=0 \\\\\\\frac{6y-y^2-(y+3+y^2+3y)-(y-y^2+5-5y)}{y(1-y)(1+y)} =0\\\\\\\frac{6y-y^2-y^2-4y-3+y^2+4y-5}{y(1-y)(1+y)} =0\\\\\\\frac{-y^2+6y-8}{y(1-y)(1+y)}=0\\\\\\\frac{-(y^2-6y+8)}{y(1-y)(1+y)}=0 \\\\\\\frac{y^2-6y+8}{y(y-1)(y+1)}=0\\\\\\[/tex]

ОДЗ: у≠0  у-1≠0   у≠1   у+1≠0   у≠-1

[tex]y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0[/tex]

y-2=0

y₁=2 ∈ОДЗ

y-4=0

y₂=4 ∈ОДЗ.