Question

Repede va rog! e tema pentru luni!

Answer 1

a) $\frac{- 1235}{197})^0$ = 1
b) $( \frac{-36}{47} )^1$ = $\frac{-36}{47}$
c) $\frac{-1997}{1997}^{1997}$ = -1
d) $\frac{-17}{34}^4$ = $-0,5^{4}$ = 0,0625
e) $( -4\frac{32}{124} )^{1997}$ = 0
f) $0^{2015} = 0$
g)  $(-\frac{23}{23 }) ^{-1} = -1^{-1} = -1$
h)$(-1,5)^{-2} = ( \frac{1}{1,5^2})= \frac{1}{-2,25}$
i) $[0,(3)]^{4}$ = $( \frac{1}{3}) ^{4}$
j) $( -1\frac{2}{3} ) ^{-2} = ( -\frac{3}{3} ) ^{-2}= ( - \frac{3}{3 ^2})=( -\frac{3}{9})$
k) $[-1,1(6)] ^{-3} = ( -\frac{105}{90 ^{2} }) =( \frac{105}{8100} )$

Answer 2

a) 1^0=1, formula a^0=1; b) -36/47, formula a¹=a. c)-1^1997=-1, formula a¹=a, (-1)ⁿ=1, (-1)^ⁿ+¹=-1 (adica daca nr negativ, nu numai -1daca are exponent par rezultatul va fi pozitiv, daca exponent impar atunci nr primit va fi cu minus . d)( -1/2)⁴=1/16. e). (-528/124)^1997= (-4 intreg 4/ 31)^1997= (128/31)^1997= 128^1997/31^1997, formula (a/b)ⁿ=aⁿ/bⁿ. f) 0, formula 0ⁿ=0. g)(-1)^-¹= 1/(-1)¹=-1, formula a^-ⁿ=1/aⁿ. h)1/(-1,5)²=1/2,25=0,(4). i) [0,(3)]⁴=(3/9)⁴=(1/3)⁴=1/81. j)1/(-5/3)²=1/(25/9)=9/25, formula (a/b)/(c/d)=(a/b)*(d/c) adica inmult cu inversul impartitorului. k) [-1,1(6)]^-³= 1/(105/90)³= 1/(7/6)³= 6³/7³= 216/343.