4y do trzeciej - 11y do drugiej przez 3x do drugiej - 18y +27 =
Wyznacz dziedzine wyrazenia wymiernego
(4y³-11y²) / (3x²-18y+27)
3x²-18y+27=0
Δ=b²-4ac=324-4*3*27=324-324=0
x₀=-b/2a=-(-18)/2*3=18/6=3
D=R /{3}
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(4y³-11y²) / (3x²-18y+27)
3x²-18y+27=0
Δ=b²-4ac=324-4*3*27=324-324=0
x₀=-b/2a=-(-18)/2*3=18/6=3
D=R /{3}