a)
\frac{2x+1}{ x^{2} +6x+9} + \frac{x-1}{9- x^{2} }=0
\frac{2x+1}{(x+3)^2}-\frac{x-1}{x^2-9}=0
\frac{2x+1}{(x+3)^2}-\frac{x-1}{(x+3)(x-3)}=0
\frac{(2x+1)(x-3)-(x-1)(x+3)}{(x+3)^2(x-3)}=0
\frac{2x^2-6x+x-3-(x^2+3x-x-3)}{(x+3)^2(x-3)}=0
\frac{2x^2-5x-3-x^2-2x+3}{(x+3)2(x-3)}=0
x^2-7x=0
x(x-7)=0
x=0\vee x=7
b)
\frac{x+5}{x-4}+ \frac{3}{x}= \frac{36}{ x^{2}-4x }
x-4\ne0=> x\ne4 i x\ne0
D=R \ {0,4}
\frac{x(x+5)+3(x-4)}{x(x-4)}-\frac{36}{x(x-4)}=0
\frac{x^2+5x+3x-12-36}{x^2-4x}=0
x^2+8x-48=0
a=1, b=8, c=-48
\Delta=b^2-4ac=8^2-4*1*(-48)=64+192=256
\sqrt\Delta=16
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-8-16}{2}=\frac{-24}{2}=-12
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{-8+16}{2}=\frac{8}{2}=4 nie należy do dziedziny
x=-12
c)
\frac{x}{x+2}- \frac{1}{2-x}= \frac{4}{ x^{2} -4}
\frac{x}{x+2}-\frac{1}{-(x-2)}=\frac{4}{x^2-4}
\frac{x}{x+2}+\frac{1}{x-2}=\frac{4}{x^2-4}
\frac{x(x-2)+x+2}{(x+2)(x-2)}-\frac{4}{x^2-4}=0
\frac{x^2-2x+x+2-4}{x^2-4}=0
\frac{x^2-x-2}{x^2-4}=0
x^2-x-2=0
a=1, b=-1, c=-2
\Delta=1-4*1*(-2)=9
\sqrt\Delta=3
x_1=\frac{1-3}{2}=\frac{-2}{2}=-1
x_2=\frac{1+3}{2}=2 nie należy do dziedziny