Zadanie 2
Oblicz wartość sumy algebraicznej dla x = -1/2 , x=-1/4, i x = 1/4
a)
4x^2-8x
1)
x=-\frac{1}{2}
4x^2-8x=4(x^2-2x)=4[(-\frac{1}{2})^2-2*(-\frac{1}{2}]=4(\frac{1}{4}+\frac{8}{2})=4(\frac{1}{4}+4)=1+16=17
2)
x=-\frac{1}{4}
4x^2-8x=4(x^2-2x)=4[(-\frac{1}{4})^2-2*\frac{1}{4}]=4(\frac{1}{16}-\frac{2}{4})=\not4^1*\frac{1}{\not16^4}-4*\frac{2}{4}=
=\frac{1}{4}-\frac{8}{4}=-\frac{7}{4}=-1\frac{3}{4}
3)
x=\frac{1}{4}
4x^2-8x=4(x^2-2x)=4[(\frac{1}{4})^2-2*(\frac{1}{4})]=4(\frac{1}{16}-\frac{2}{4})=\frac{4}{16}-\frac{8}{4}=\frac{1}{4}-\frac{8}{4}=-\frac{7}{4}=-1\frac{3}{4}
b)
1)
x=-\frac{1}{2}
-8x^3+2x=-2(4x^3-x)=-2*[4*(-\frac{1}{2})^3-(-\frac{1}{2}]=-2[\not4^1*(-\frac{1}{\not8^2})+\frac{1}{2})=
=-2(-\frac{1}{2}+\frac{1}{2})=-2*0=0
2)
x=-\frac{1}{4}
-8x^3+2x=-2(4x^3-x)=-2[4*(-\frac{1}{4})^3-2*(-\frac{1}{4})]=-2[-\not 4^1*\frac{1}{\not64^{16}}+\frac{1}{4}=
=-2(-\frac{1}{16}+\frac{1}{4})=\frac{2}{16}-\frac{2}{4}=\frac{1}{8}-\frac{4}{8}=-\frac{3}{8}
3)
x=\frac{1}{4}
-8x^3+2x=-2(4x^3-x)=-2[4*(\frac{1}{4})^3-\frac{1}{4}]=-2(\not4^1*\frac{1}{\not64^{16}}-\frac{1}{4})=-2(\frac{1}{16}-\frac{1}{4})=
=-\frac{2}{16}+\frac{2}{4}=-\frac{1}{8}+\frac{4}{8}=\frac{3}{8}
c)
1)
x=-\frac{1}{2}
16x^4-x^2=16*(-\frac{1}{2})^4-(-\frac{1}{2})^2=\not16^1*\frac{1}{\not16^1}-\frac{1}{4}=1-\frac{1}{16}=\frac{4}{4}-\frac{1}{4}=\frac{3}{4}
2)
x=-\frac{1}{4}
16x^4-x^2=16*(-\frac{1}{4})^4-(-\frac{1}{4})^2=\not16^1*\frac{1}{\not256^{16}}-\frac{1}{16}=\frac{1}{16}-\frac{1}{16}=0
Liczba ujemna do potęgi parzystej daje liczbę dodatnią.
3)
x=\frac{1}{4}
16x^4-x^2=16*(\frac{1}{4})^4-(\frac{1}{4})^2=\not16^1*\frac{1}{\not256^{16}}-\frac{1}{16}=\frac{1}{16}-\frac{1}{16}=0