a)
\frac{3x-2}{2x+1}= -2
2x+1\ne0 => 2x\ne 1 => x\ne \frac{1}{2} => D=R \ {1/2}
3x-2=-2(2x+1)
3x-2=-4x-2
3x+4x=-2+2
7x=0/:7
x=0
b)
\frac{3x+4}{x+2}= \frac{x+8}{x+5}
x+2\ne0 i x+5\ne0 => x\ne -2 i x\ne -5 => D=R \ {-5,-2}
(3x+4)(x+5)=(x+8)(x+2)
3x^2+15x+4x+20=x^2+2x+8x+16
3x^2+19x+20=x^2+10x+16
3x^2+19x+20-x^2-10x-16=0
2x^2+9x+4=0
a=2, b=9, c=4
\Delta=b^2-4ac=81-4*2*4=81-32=49
\sqrt\Delta=7
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-9-7}{4}=\frac{-16}{4}=-4
x_2=\frac{-b-\sqrt\Delta}{2a}=\frac{-9+7}{4}=\frac{-2}{4}=-\frac{1}{2}
c)
\frac{-3x+4}{x-1} =\frac{4}{x-4} -3
x-1\ne 0 i x-4\ne 0 => x\ne 1 i x\ne 4 => D=R \ {1,4)
\frac{-3x+4}{x-1}=\frac{4-3(x-4)}{x-4}
\frac{-3x+4}{x-1}=\frac{4-3x+12}{x-4}
\frac{-3x+4}{x-1}=\frac{16-3x}{x-4}
(-3x+4)(x-4)=(16-3x)(x-1)
-3x^2+12x+4x-16=16x-16-3x^2+3x
-3x^2+16x-16=-3x^2+19x-16
-3x^2+16x-16+3x^2-19x+16=0
-3x=0/:3
x=0