Question

Va implor ajutor nu știu să rezolv!!!HELP!
Exercițiul 2 punctul b,c!

Answer 1

I : $/2^{51} - 3^{34}/ = / ( 2^{3})^{17} / =\ \textgreater \ numar negativ =\ \textgreater \ -(2^{51}-3^{34}$
$= -2^{51}+3^{34}$
$deoarece : 2^{3}\ \textless \ 3^{2} =\ \textgreater \ (2^{2})^{17}$
$=[2^{51}-2^{51}+3^{34}] : [-(3^{2})^{16})] =3^{34} : (-3^{22})$
$=3^{34} : (-3^{22})=-3^{34}-^{32}=-3^{2}=-9$

Answer 2

b)

[tex]\it (-3)^0+(-3)^{42}\cdot(-3^{25}):81^{16} =1+3^{42}\cdot(-3)^{25} :(3^4)^{16}= \\\;\\ 1-3^{42}\cdot3^{25}:3^{64} =1-3^{42+25-64} =1-3^3=1-27=-26[/tex]

c)

[tex]\it 2^{51} =2^{3\cdot17} =(2^3)^{17} =8^{17} \\\;\\ 3^{34}=3^{2\cdot17} =(3^2)^{17} =9^{17}[/tex]

[tex]\it8^{17}\ \textless \ 9^{17} \Rightarrow 2^{51}\ \textless \ 3^{34} \Rightarrow 2^{51}-3^{34} \ \textless \ 0\Rightarrow |2^{51}-3^{34} | = \\\;\\ =-(2^{51}+3^{34}) =-2^{51}+3^{34} [/tex]

Acum, exercițiul devine:

[tex]\it (\not2^{51}-\not2^{51}+3^{34}):(-9^{16}) = (3^2)^{17}:(-9^{16}) = 9^{17}:(-9^{16}) = \\\;\\ -9^{17}:9^{16} = -9^{17-16} =-9^1=-9[/tex]