Verified answer

Відповідь:

a) [tex]x_{1} =-5 , x_{2} =3[/tex]    б)[tex]x_{1} =-15,x_{2} =1[/tex]

Пояснення:

а)[tex]x\neq -1 , x\neq 1[/tex]

[tex]\frac{12}{x+1} +\frac{8}{1-x^{2} } +\frac{x-9}{x-1} +1=0[/tex]

[tex]\frac{12}{x+1} - \frac{8}{(x-1)(x+1)} +\frac{x-9}{x-1} +1=0[/tex]

[tex]\frac{12x-12-8+x^{2} -9x+x-9+x^{2} -1}{(x-1)(x+1)}[/tex]

[tex]\frac{4x-30+2x^{2} }{(x-1)(x+1)} =0[/tex]

[tex]4x-30+2x^{2} =0[/tex]

[tex]2x^{2} +4x-30=0[/tex]

[tex]x^{2} +5x-3x-15=0\\ x(x+5)-3(x+5)=0\\ (x+5)*(x-3)=0[/tex]

[tex]x+5=0,x=-5,\\ x-3=0, x=3, x\neq -1,x\neq 1[/tex]           [tex]x=-5,x=3[/tex]

б)[tex]x\neq 9,x\neq 0,x\neq -9.[/tex]

[tex]\frac{3}{x-9} + \frac{5}{x(x+9)} +\frac{10}{(x-9)(9+x)} =0\\\frac{3x^{2} +42x-45}{x(x-9)(x+9)} =0\\3x^{2} +42x-45=0\\x^{2} +14x-15=0\\x^{2} +15x-x-15=0\\x(x+15)-(x-15)=0\\(x+15)*(x-1)=0\\x+15=0,x=-15\\x-1=0,x=1.[/tex]