Question

calculati:a)9^0+1^8+8^1+0^2004;b)7*2^1997-6*2^1996-2*2^1998 va rog pt 20 pcte

Answer 1

a)
${9}^{0} + {1}^{8} + {8}^{1} + {0}^{2004} = \\ 1 + 1 + 8 + 0 = 10$
Deoarece orice numa la puterea 0 e ega cu 1, si 0 la orice putere e 0

b)
$7 \times {2}^{1997} - 6 \times {2}^{1996} - 2 \times {2}^{1998} = \\ {2}^{1996} (7 \times 2 - 6 - 2 \times {2}^{2} ) = \\ {2}^{1996} \times (14 - 14) = \\ {2}^{1996} \times 0 = 0$

Answer 2

9^0=1, 1^8=1, 8^1=8, 0^2004=0