Liczba kjest rozwiązaniem równania x√5 = x + 2, a liczba m jest rozwiązaniem równania (2-x)² - √5 = (x-1) (x-5). Między liczbami k i m zachodzi zależność:
A. k = m
B. k > m
C. k < m
D. k · m = 0
Proszę o wyjaśnienie.
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Obliczam m
![x\sqrt{5}=x+2 x\sqrt{5}=x+2](https://tex.z-dn.net/?f=x%5Csqrt%7B5%7D%3Dx%2B2)
![x\sqrt{5}-x=2 x\sqrt{5}-x=2](https://tex.z-dn.net/?f=x%5Csqrt%7B5%7D-x%3D2)
![x(\sqrt{5}-1)=2 x(\sqrt{5}-1)=2](https://tex.z-dn.net/?f=x%28%5Csqrt%7B5%7D-1%29%3D2)
![x= \frac{2}{\sqrt{5}-1} x= \frac{2}{\sqrt{5}-1}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B2%7D%7B%5Csqrt%7B5%7D-1%7D)
![x= \frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} x= \frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B2%28%5Csqrt%7B5%7D%2B1%29%7D%7B%28%5Csqrt%7B5%7D-1%29%28%5Csqrt%7B5%7D%2B1%29%7D)
![x= \frac{2(\sqrt{5}+1)}{5-1} x= \frac{2(\sqrt{5}+1)}{5-1}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B2%28%5Csqrt%7B5%7D%2B1%29%7D%7B5-1%7D)
![x= \frac{2(\sqrt{5}+1)}{4} x= \frac{2(\sqrt{5}+1)}{4}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7B2%28%5Csqrt%7B5%7D%2B1%29%7D%7B4%7D)
![x=\frac{\sqrt{5}+1}{2} x=\frac{\sqrt{5}+1}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B2%7D)
![(2-x)^2-\sqrt{5}=(x-1)(x-5) (2-x)^2-\sqrt{5}=(x-1)(x-5)](https://tex.z-dn.net/?f=%282-x%29%5E2-%5Csqrt%7B5%7D%3D%28x-1%29%28x-5%29)
![4-4x+x^2-\sqrt{5}=x^2-5x-x+5 4-4x+x^2-\sqrt{5}=x^2-5x-x+5](https://tex.z-dn.net/?f=4-4x%2Bx%5E2-%5Csqrt%7B5%7D%3Dx%5E2-5x-x%2B5)
![-4x+x^2-x^2+5x+x=5+\sqrt{5}-4 -4x+x^2-x^2+5x+x=5+\sqrt{5}-4](https://tex.z-dn.net/?f=-4x%2Bx%5E2-x%5E2%2B5x%2Bx%3D5%2B%5Csqrt%7B5%7D-4)
![2x= \sqrt{5}+1 2x= \sqrt{5}+1](https://tex.z-dn.net/?f=2x%3D+%5Csqrt%7B5%7D%2B1)
![x=\frac{\sqrt{5}+1}{2} x=\frac{\sqrt{5}+1}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Csqrt%7B5%7D%2B1%7D%7B2%7D)
![m=k m=k](https://tex.z-dn.net/?f=m%3Dk)
Obliczam k