Question

ex 2 va roggg imi trebuie urgent (sa.mi scrieti si rezolvarea )

Answer 1

$\displaystyle 2a).(-7)^1=-7 \\ b).(-7)^1+(-7)^0=-7+1=-6 \\ c).-7^0+(-7)^0=-1+1=0 \\ d).1003^0=1 \\ e).4+(-1)^1=4+(-1)=4-1=3 \\ f).(-129)^1-(-5)^3=-129-(-125)=-129+125=-4 \\ g) .-29^0+(-103)^0=-1+1=0 \\ h).-2^2+(+1)^5=-4+1=-3 \\ i).-13^0+13^1+(-13)^1+1^0+1999^0=-1+13-13+1+1= \\ =12-13+1+1=-1+1+1=0+1=1 \\ j).(-2)^3+(+3)^2=-8+9=1 \\ k).-6^2+(-5)^2=-36+25=-11$
$\displaystyle l).[(-10)^4]^5:(-10)^{18}-3^4=(-10)^{20}:(-10)^{18}-3^4= \\ =(-10)^{20-18}-81=(-10)^2-81=100-81=19 \\ m).7:23^0=7:1=7 \\ n).11^5 \cdot 11^{103}-(+11)^{108}=11^{5+103}-11^{108}=11^{108}-11^{108}=0 \\ o).(-7)^{2011}:(-7)^{2009}=(-7)^{2011-2009}=(-7)^2=49 \\ p).[(-3)^4]^{60}:(-3)^{238}+10^2=(-3)^{240}:(-3)^{238}+100= \\ =(-3)^{240-238}+100=(-3)^2+100=9+100=109 \\ r).79^{43}:79^{34}-7799^0 \cdot (79^3)^3=79^{43-34}-1 \cdot 79^9= \\ =79^9-1 \cdot 79^9=79^9-79^9=0$