Odpowiedź :
Odpowiedź:
x⁴+3x³+4x²= x²(x²+3x+4)
Δ= 3²-4*1*4= -7 Δ<0 = koniec zadania
3]
x-1≠0 x≠1
(x+1)(x-1)= 2-2x x²-1+2x-2=0 x²+2x-3=0
Δ=b²-4ac=4+12=16 √Δ=4
x1=(-b-√Δ)/2a=( -2-4)/2=-3
x2=(-b+√Δ)/2a=(-2+4)/2= 1= sprzeczne ∉ D
odp. x= -3
Szczegółowe wyjaśnienie:
2.
[tex]W(x) = x^{4}+3x^{3}+4x^{2} = x^{2}(x^{2}+3x+4)[/tex]
3.
[tex]x+1 = \frac{2-2x}{x-1} \ \ /\cdot(x-1)\\\\\\Z:\\x-1 \neq 0\\x \neq 1\\D = R \setminus\{1\}\\\\\\(x+1)(x-1) = 2-2x\\\\x^{2}-1 = 2-2x\\\\x^{2}+2x-1-2 = 0\\\\x^{2}+2x-3 = 0\\\\a = 1, \ b = 2, \ c = -3\\\\\Delta = b^{2}-4ac = 2^{2}-4\cdot1\cdot(-3) = 4+12 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-2-4}{2\cdot1}=\frac{-6}{2} = -3\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-2+4}{2} = \frac{2}{2} = 1 \ \notin D\\\\\boxed{x = -3}[/tex]