Odpowiedź:
Szczegółowe wyjaśnienie:
a.
[tex]x^2-4x+5=0[/tex]
Δ=16-4*1*5=-4
Δ<0 , brak rozwiązań
b.
[tex]x^2-x-1=0\\[/tex]
Δ=1-4*1*(-1)=1+4=5, √Δ=√5
[tex]x_1=\frac{1-\sqrt{5}} {2}[/tex] [tex]x_2=\frac{1+\sqrt{5}} {2}[/tex]
c.
[tex]x^2-x+2=0[/tex]
Δ=1-4*1*2=1-8=-7
Δ<0, brak rozwiązań
d.
[tex]\frac{2x}{x^2-1} =\frac{1}{x-1} +\frac{1}{x+1} +1[/tex]
D:
[tex]x^2-1\neq 0\\(x-1)(x+1)\neq 0[/tex]
[tex]x_1\neq 1[/tex] [tex]x_2\neq -1[/tex]
D=R\{-1, 1}
[tex]\frac{2x}{(x-1)(x+1)} =\frac{1}{x-1} +\frac{1}{x+1} +1[/tex] /*(x-1)(x+1)
[tex]2x=(x+1)+(x-1)+(x-1)(x+1)\\2x=2x+x^2-1\\x^2-1=0\\(x-1)(x+1)=0\\[/tex]
[tex]x_1=1[/tex] ∉D [tex]x_2=-1[/tex] ∉D
brak rozwiązań
e.
[tex]\frac{2x-6}{x^2-11x+30} +\frac{7x-2}{x^2-2x-24} =\frac{8x+5}{x^2-x-20}[/tex]
D: [tex]x^2-11x+30\neq 0[/tex] ∨ [tex]x^2-2x-24\neq 0[/tex] ∨ [tex]x^2-x-20\neq 0[/tex]
Δ=121-4*1*30=1 Δ=4-4*1*(-24)=100 Δ=1-4*1*(-20)=81
√Δ=1 √Δ=10 √Δ=9
[tex]x_1\neq 5[/tex] v [tex]x_2\neq 6[/tex] v [tex]x_3\neq -4[/tex] v [tex]x_4\neq 6[/tex] v [tex]x_5\neq -4[/tex] v [tex]x_5\neq 5[/tex]
D=R\{-4, 5, 6}
[tex]\frac{2x-6}{(x-5)(x-6)} +\frac{7x-2}{(x+4)(x-6)} =\frac{8x+5}{(x+4)(x-5)}[/tex] /*(x-5)(x-6)(x+4)
[tex](x+4)(2x-6)+(x-5)(7x-2)=(x-6)(8x+5)\\2x^2-6x+8x-24+7x^2-2x-35x+10=8x^2+5x-48x-30\\x^2+8x+16=0[/tex]
Δ=64-4*1*16=0
[tex]x=\frac{-b}{2a}=\frac{-8}{2} =-4[/tex] ∉D
[tex]x^2-4x+5=0\\a=1, b=-4, c=5\\\Delta=(-4)^2-4*1*5=16-20=-4\\\Delta < 0[/tex]
Brak rozwiązań.
[tex]x^2-x-1=0\\a=1, b=-1, c=-1\\\Delta=(-1)^2-4*1*(-1)=1+4=5\\\sqrt{\Delta}=5\\\boxed{x_{1, 2}=\dfrac{1\pm\sqrt5}2}[/tex]
[tex]x^2-x+2=0\\a=1, b=-1, c=2\\\Delta=(-1)^2-4*1*2=1-8=-7\\\Delta < 0[/tex]
[tex]\dfrac{2x}{x^2-1}=\dfrac{1}{x-1}+\dfrac{1}{x+1}+1\\\\(x^2-1)(x-1)(x+1) \neq 0\\(x-1)(x+1)(x-1)(x+1) \neq 0\\(x-1)^2(x+1)^2 \neq 0\\x\neq 1 \wedge x \neq -1\\\\\underline{D: x\in \mathbb{R} \setminus \{-1, 1\}}[/tex]
[tex]\dfrac{2x}{(x-1)(x+1)}=\dfrac{x+1}{(x-1)(x+1)}+\dfrac{x-1}{(x-1)(x+1)}+\dfrac{(x+1)(x-1)}{(x+1)(x-1)}\\2x=x+1+x-1+x^2-1\\2x=2x+x^2-1\\0=x^2-1\\(x-1)(x+1)=0\\x =1 \notin D, x=-1 \notin D[/tex]
[tex]\dfrac{2x-6}{x^2-11x+30}+\dfrac{7x-2}{x^2-2x-24}=\dfrac{8x+5}{x^2-x-20}[/tex]
[tex](x^2-11x+30)(x^2-2x-24)(x^2-x-20)\neq 0[/tex]
[tex]\begin{array}{ccccc}x^2-11x+30 \neq 0 & \wedge & x^2-2x-24\neq 0 & \wedge & x^2-x-20 \neq 0\\\Delta=1&&\Delta=100&&\Delta=81\\\sqrt{\Delta}=1&&\sqrt{\Delta}=10&&\sqrt{\Delta}=9\\x_{1,2}\neq\dfrac{11\pm1}{2}&\wedge&x_{3,4}\neq \dfrac{2\pm10}{2}&\wedge&x_{5,6}\neq\dfrac{1\pm9}{2}\\x\neq6 \wedge x\neq5&\wedge&x\neq6\wedge x\neq -4 & \wedge & x\neq 5 \wedge x\neq-4\end{array}[/tex]
[tex]x\neq -4 \wedge x \neq 5 \wedge x\neq 6\\\\\underline{D: x\in \mathbb{R} \setminus\{-4, 5, 6\}}[/tex]
[tex]\dfrac{2x-6}{(x-6)(x-5)}+\dfrac{7x-2}{(x-6)(x+4)}=\dfrac{8x+5}{(x-5)(x+4)} /*(x-6)(x-5)(x+4)\\\\(x+4)(2x-6)+(x-5)(7x-2)=(x-6)(8x+5)\\\\2x^2+2x-24+7x^2-37x+10=8x^2-43x-30\\\\x^2+8x+16=0\\a=1, b=8, c=16\\\\\Delta=8^2-4*1*16=64-64=0\\\\x_0=\dfrac{-8}2=-4 \notin D[/tex]
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Odpowiedź:
Szczegółowe wyjaśnienie:
a.
[tex]x^2-4x+5=0[/tex]
Δ=16-4*1*5=-4
Δ<0 , brak rozwiązań
b.
[tex]x^2-x-1=0\\[/tex]
Δ=1-4*1*(-1)=1+4=5, √Δ=√5
[tex]x_1=\frac{1-\sqrt{5}} {2}[/tex] [tex]x_2=\frac{1+\sqrt{5}} {2}[/tex]
c.
[tex]x^2-x+2=0[/tex]
Δ=1-4*1*2=1-8=-7
Δ<0, brak rozwiązań
d.
[tex]\frac{2x}{x^2-1} =\frac{1}{x-1} +\frac{1}{x+1} +1[/tex]
D:
[tex]x^2-1\neq 0\\(x-1)(x+1)\neq 0[/tex]
[tex]x_1\neq 1[/tex] [tex]x_2\neq -1[/tex]
D=R\{-1, 1}
[tex]\frac{2x}{(x-1)(x+1)} =\frac{1}{x-1} +\frac{1}{x+1} +1[/tex] /*(x-1)(x+1)
[tex]2x=(x+1)+(x-1)+(x-1)(x+1)\\2x=2x+x^2-1\\x^2-1=0\\(x-1)(x+1)=0\\[/tex]
[tex]x_1=1[/tex] ∉D [tex]x_2=-1[/tex] ∉D
brak rozwiązań
e.
[tex]\frac{2x-6}{x^2-11x+30} +\frac{7x-2}{x^2-2x-24} =\frac{8x+5}{x^2-x-20}[/tex]
D: [tex]x^2-11x+30\neq 0[/tex] ∨ [tex]x^2-2x-24\neq 0[/tex] ∨ [tex]x^2-x-20\neq 0[/tex]
Δ=121-4*1*30=1 Δ=4-4*1*(-24)=100 Δ=1-4*1*(-20)=81
√Δ=1 √Δ=10 √Δ=9
[tex]x_1\neq 5[/tex] v [tex]x_2\neq 6[/tex] v [tex]x_3\neq -4[/tex] v [tex]x_4\neq 6[/tex] v [tex]x_5\neq -4[/tex] v [tex]x_5\neq 5[/tex]
D=R\{-4, 5, 6}
[tex]\frac{2x-6}{(x-5)(x-6)} +\frac{7x-2}{(x+4)(x-6)} =\frac{8x+5}{(x+4)(x-5)}[/tex] /*(x-5)(x-6)(x+4)
[tex](x+4)(2x-6)+(x-5)(7x-2)=(x-6)(8x+5)\\2x^2-6x+8x-24+7x^2-2x-35x+10=8x^2+5x-48x-30\\x^2+8x+16=0[/tex]
Δ=64-4*1*16=0
[tex]x=\frac{-b}{2a}=\frac{-8}{2} =-4[/tex] ∉D
brak rozwiązań
Verified answer
a)
[tex]x^2-4x+5=0\\a=1, b=-4, c=5\\\Delta=(-4)^2-4*1*5=16-20=-4\\\Delta < 0[/tex]
Brak rozwiązań.
b)
[tex]x^2-x-1=0\\a=1, b=-1, c=-1\\\Delta=(-1)^2-4*1*(-1)=1+4=5\\\sqrt{\Delta}=5\\\boxed{x_{1, 2}=\dfrac{1\pm\sqrt5}2}[/tex]
c)
[tex]x^2-x+2=0\\a=1, b=-1, c=2\\\Delta=(-1)^2-4*1*2=1-8=-7\\\Delta < 0[/tex]
Brak rozwiązań.
d)
[tex]\dfrac{2x}{x^2-1}=\dfrac{1}{x-1}+\dfrac{1}{x+1}+1\\\\(x^2-1)(x-1)(x+1) \neq 0\\(x-1)(x+1)(x-1)(x+1) \neq 0\\(x-1)^2(x+1)^2 \neq 0\\x\neq 1 \wedge x \neq -1\\\\\underline{D: x\in \mathbb{R} \setminus \{-1, 1\}}[/tex]
[tex]\dfrac{2x}{(x-1)(x+1)}=\dfrac{x+1}{(x-1)(x+1)}+\dfrac{x-1}{(x-1)(x+1)}+\dfrac{(x+1)(x-1)}{(x+1)(x-1)}\\2x=x+1+x-1+x^2-1\\2x=2x+x^2-1\\0=x^2-1\\(x-1)(x+1)=0\\x =1 \notin D, x=-1 \notin D[/tex]
Brak rozwiązań.
e)
[tex]\dfrac{2x-6}{x^2-11x+30}+\dfrac{7x-2}{x^2-2x-24}=\dfrac{8x+5}{x^2-x-20}[/tex]
[tex](x^2-11x+30)(x^2-2x-24)(x^2-x-20)\neq 0[/tex]
[tex]\begin{array}{ccccc}x^2-11x+30 \neq 0 & \wedge & x^2-2x-24\neq 0 & \wedge & x^2-x-20 \neq 0\\\Delta=1&&\Delta=100&&\Delta=81\\\sqrt{\Delta}=1&&\sqrt{\Delta}=10&&\sqrt{\Delta}=9\\x_{1,2}\neq\dfrac{11\pm1}{2}&\wedge&x_{3,4}\neq \dfrac{2\pm10}{2}&\wedge&x_{5,6}\neq\dfrac{1\pm9}{2}\\x\neq6 \wedge x\neq5&\wedge&x\neq6\wedge x\neq -4 & \wedge & x\neq 5 \wedge x\neq-4\end{array}[/tex]
[tex]x\neq -4 \wedge x \neq 5 \wedge x\neq 6\\\\\underline{D: x\in \mathbb{R} \setminus\{-4, 5, 6\}}[/tex]
[tex]\dfrac{2x-6}{(x-6)(x-5)}+\dfrac{7x-2}{(x-6)(x+4)}=\dfrac{8x+5}{(x-5)(x+4)} /*(x-6)(x-5)(x+4)\\\\(x+4)(2x-6)+(x-5)(7x-2)=(x-6)(8x+5)\\\\2x^2+2x-24+7x^2-37x+10=8x^2-43x-30\\\\x^2+8x+16=0\\a=1, b=8, c=16\\\\\Delta=8^2-4*1*16=64-64=0\\\\x_0=\dfrac{-8}2=-4 \notin D[/tex]
Brak rozwiązań.