1Nomor 1PembahasanDiketahui a=12a = \frac{1}{2}a=21, b=2b = 2b=2, c=1c = 1c=1 Nilai dari a−2bc3ab2c−1=....\frac{a^{-2}bc^3}{ab^2c^{-1}} = ....ab2c−1a−2bc3=....111222333444555PembahasanDiketahui: a=12a = \frac{1}{2}a=21; b=2b = 2b=2, c=1c = 1c=1 a−2bc3ab2c−1=bc3ab2⋅c1a2\frac{a^{-2}bc^3}{ab^2c^{-1}} = \frac{bc^3}{ab^2} \cdot \frac{c^1}{a^2}ab2c−1a−2bc3=ab2bc3⋅a2c1=2⋅13(12)⋅22⋅11(12)2= \frac{2 \cdot 1^3}{\left(\frac{1}{2}\right) \cdot 2^2} \cdot \frac{1^1}{\left(\frac{1}{2}\right)^2}=(21)⋅222⋅13⋅(21)211=2⋅42=4= \frac{2 \cdot 4}{2} = 4=22⋅4=4
