\frac{4x-3}{x}= \frac{3(x+1)}{x^2-x}
dziedzina
x\ne 0
i
x^2-x\ne 0
x(x-1)\ne0
x\ne 0 i x\ne 1 , d=\mathbb R\backslash \{0,1\}
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\frac{4x-3}{x}= \frac{3(x+1)}{x^2-x}
(4x-3)(x^2-x)=3(x+1)*x
4x^3-4x^2-3x^2+3x=3x^2+3x
4x^3-7x^2+3x-3x^2-3x=0
4x^3-10x^2=0
2x^2(2x-5)=0
x^2=0\to x=0 \not\in D
lub
2x-5=0
2x=5
x=\frac{5}{2}
x=2,5