a)
\frac{3x-4}{2} = \frac{4x-5}{3} mnożę “na krzyż”
3(3x-4)=2(4x-5)
9x-12=8x-10 |-8x od obu stron równania
x-12=-10 |+12 do obu stron równania
x=2
b)
\frac{x}{2} = \frac{8}{x}
x\ne 0 , D = \mathbb R - {0}
x*x=2*8
x^2=16
x^2=4^2
x=4
c)
\frac{2}{x^2} = \frac{1}{3}
x^2\ne 0 => x\ne 0 , D = \mathbb R - {0}
x^2=2*3
x^2=6
x_1=-\sqrt6 , x_2=\sqrt6
x \in {-\sqrt6, \sqrt6}