Question

5)Calculați,aplicând regulile de calcul cu puteri:
a) ($-\frac{1}{2}$)³ ×($-\frac{1}{2}$)⁵=
b)($\frac{-5}{-3}$)⁰×($\frac{-5}{-3}$)⁶=
c)(0,1)²×(0,1)³×(0,1)⁴×(0,1)⁰=
d)($-\frac{1}{5}$⁶:($-\frac{1}{5}$)⁴=
e)($-\frac{3}{8}$)⁵:($-\frac{3}{8}$)⁴=
f)(0.2)⁷>(0.2)⁵=
g)(1,5)¹⁹:(1,5)¹⁷=
h)[($\frac{3}{2}$)²]³=
i)[($-\frac{3}{8}$)²]³=
j)[($\frac{1}{-2}$)³]²=
k){[(-0.1)²]³}⁴=

VERIFICARE:

a)=($-\frac{1}{2}$)⁸
b)=($\frac{5}{3}$)⁶
c)=(0.1)⁹
d)=($-\frac{1}{5}$)²
e)=$-\frac{3}{8}$
f)=(0.2)²
g)=(1.5)²
h)($\frac{3}{2}$)⁶
i)($\frac{3}{8}$⁶
j)($\frac{1}{2}$)⁶
k)(0.1)²⁴.

Answer 1

(-1/2)^8
b) (-5/-3)^6=(5/3)^6
c)=(0,1)^(2+3+4+0)=(0,1)^9
d) =(-1/5)^²=(1/5)²
e) =(-3/8)
f)=(0,2)²
g)=(1,5)²
h) =(3/2)^2.3=(3/2)^6
i) =(-3/8)^6=(3/8)^6
j)=(1/2)^6
k)(-0,1)^2.3.4=(-0,1)^24=(0,1)^24
se bazeaza pe formule la inmultire se aduna puterile,la impartire se scad ,putere la alta putere se inmultesc puterile .un numar negativ la o putere cu sot da + si la una fara sot da -