Zadanie 2
dla : x=-½, x=-¼ i x=¼
a) 4x²-8x
4\cdot (-\frac{1}{2})^2-8\cdot (-\frac{1}{2})=\not4\cdot \frac{1}{\not4}+4=1+4=5
4\cdot (-\frac{1}{4})^2-\not8^2\cdot (-\frac{1}{\not4^1})=\not4^1\cdot \frac{1}{\not16^4}-2\cdot (-1)=\frac{1}{4}+2=2\frac{1}{4}
4\cdot (\frac{1}{4})^2-\not8^2\cdot (\frac{1}{\not4^1})=\not4^1\cdot \frac{1}{\not16^4}-2\cdot 1=\frac{1}{4}-2=-(2-\frac{1}{4})=-1\frac{3}{4}
b)
-8x³+2x
-8\cdot (-\frac{1}{2})^3+\not2^1\cdot (-\frac{1}{\not2^1})=-\not8^1\cdot (-\frac{1}{\not8^1})-1=(-1)\cdot (-1)-1=1-1=0
-8\cdot (-\frac{1}{4})^3+\not2^1\cdot (-\frac{1}{\not4^2})=-\not8^1\cdot (-\frac{1}{\not64^8})-\frac{1}{2}=\frac{1}{8}-\frac{4}{8}=-\frac{3}{8}
-8\cdot (\frac{1}{4})^3+\not2^1\cdot (\frac{1}{\not4^2})=-\not8^1\cdot \frac{1}{\not64^8}+\frac{1}{2}=-\frac{1}{8}+\frac{4}{8}=\frac{3}{8}
c)
16x⁴-x²
16\cdot (-\frac{1}{2})^4-(-\frac{1}{2})^2=\not16^1\cdot \frac{1}{\not16^1}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}
16\cdot (-\frac{1}{4})^4-(-\frac{1}{4})^2=\not16^1\cdot \frac{1}{\not256^{16}}-\frac{1}{16}=\frac{1}{16}-\frac{1}{16}=0
16\cdot (\frac{1}{4})^4-(\frac{1}{4})^2=\not16^1\cdot \frac{1}{\not256^{16}}-\frac{1}{16}=\frac{1}{16}-\frac{1}{16}=0