ada yg bisa bantu? ...... yang jelas ya caranya

(a) -2 × -1 × 0 × 1 = 0 = (1/2)

[tex]( \frac{1}{2})^{-2}= \dfrac{1^{-2}}{2^{-2}}=\dfrac{1}{\frac {1}{2^2}}=\dfrac{1}{ \frac{1}{4} }=1\times\frac{4}{1}=4 \\ \\ ( \frac{1}{2})^{-1}=\dfrac{1^{-1}}{2^{-1}}=\dfrac{1}{\frac{1}{2^1}}=\dfrac{1}{ \frac{1}{2} }=1\times\frac{2}{1}=2 [/tex]

[tex]( \frac{1}{2})^0=\dfrac{1^0}{2^0}= \frac{1}{1}=1 \\ ( \frac{1}{2})^1=\frac{1}{2} [/tex]

Jadi,

a) [tex]( \frac{1}{2})^{-2}+( \frac{1}{2})^{-1}+ ( \frac{1}{2})^{0}+( \frac{1}{2})^{1}=4+2+1+ \frac{1}{2}=7\frac{1}{2} [/tex]

Soal b) caranya seperti soal yang a)

b) [tex] (\frac{1}{8})^{-2}+(\frac{1}{8})^{-1} +(\frac{1}{8})^{0}+(\frac{1}{8})^{1}=64+8+1+\frac{1}{8}=73\frac{1}{8}[/tex]