Proszę o pomoc w dwóch pierwszych przykładach z załącznika.
1.
2 - 3x/(x+2) + x/(x-1) = [2(x+2)(x-1) - 3x(x-1) + x(x+2)]/(x+2)(x-1)] =
= [2(x²-x+2x-2) - (3x²²-3x) +x²+2x]/(x²-x+2x-2) =
= (2x²+2x=4-3x²+3x+x²+2x)/(x²+x-2) = (7x-4)/(x²+x-2)
Z: x+2 ≠ 0
x ≠ -2
x-1 ≠ 0
x ≠ 1
D ∈ R \ {-2; 1}
2.
(x-4)/(x²-x) : (x²-16)/x = (x-4)/x(2x-1) * x/[(x+4)(x-4)] = 1/[(2x-1)(x+4)] = 1/(2x²+8x-x-4) = 1/(2x²+7x-4)
Z: x ≠ 0
2x-1 ≠ 0
x ≠ 1/2
x+4 ≠ 0
x ≠ -4
x-4 ≠ 0
x ≠ 4
D ∈ R \ {-4; 0; ½; 4}
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1.
2 - 3x/(x+2) + x/(x-1) = [2(x+2)(x-1) - 3x(x-1) + x(x+2)]/(x+2)(x-1)] =
= [2(x²-x+2x-2) - (3x²²-3x) +x²+2x]/(x²-x+2x-2) =
= (2x²+2x=4-3x²+3x+x²+2x)/(x²+x-2) = (7x-4)/(x²+x-2)
Z: x+2 ≠ 0
x ≠ -2
x-1 ≠ 0
x ≠ 1
D ∈ R \ {-2; 1}
2.
(x-4)/(x²-x) : (x²-16)/x = (x-4)/x(2x-1) * x/[(x+4)(x-4)] = 1/[(2x-1)(x+4)] = 1/(2x²+8x-x-4) = 1/(2x²+7x-4)
Z: x ≠ 0
2x-1 ≠ 0
x ≠ 1/2
x+4 ≠ 0
x ≠ -4
x-4 ≠ 0
x ≠ 4
D ∈ R \ {-4; 0; ½; 4}