Matematică
nastasedenisa
8

5)Calculați,aplicând regulile de calcul cu puteri: a) ([latex] -\frac{1}{2} [/latex])³ ×([latex] -\frac{1}{2} [/latex])⁵= b)([latex] \frac{-5}{-3} [/latex])⁰×([latex] \frac{-5}{-3} [/latex])⁶= c)(0,1)²×(0,1)³×(0,1)⁴×(0,1)⁰= d)([latex] -\frac{1}{5} [/latex]⁶:([latex] -\frac{1}{5} [/latex])⁴= e)([latex] -\frac{3}{8} [/latex])⁵:([latex] -\frac{3}{8} [/latex])⁴= f)(0.2)⁷>(0.2)⁵= g)(1,5)¹⁹:(1,5)¹⁷= h)[([latex] \frac{3}{2} [/latex])²]³= i)[([latex] -\frac{3}{8} [/latex])²]³= j)[([latex] \frac{1}{-2} [/latex])³]²= k){[(-0.1)²]³}⁴= VERIFICARE: a)=([latex] -\frac{1}{2} [/latex])⁸ b)=([latex] \frac{5}{3} [/latex])⁶ c)=(0.1)⁹ d)=([latex] -\frac{1}{5} [/latex])² e)=[latex] -\frac{3}{8} [/latex] f)=(0.2)² g)=(1.5)² h)([latex] \frac{3}{2} [/latex])⁶ i)([latex] \frac{3}{8} [/latex]⁶ j)([latex] \frac{1}{2} [/latex])⁶ k)(0.1)²⁴.

+0
(1) Răspunsuri
bemceva

(-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 -

Adaugă răspuns