Question

VA ROG AJUTATI-MA!!
EX5
99 DE PUNCTE+COROANA

Answer 1

$\displaystyle 5a).(-2)^{-1}+(-2)^{-2}+(-2)^{-3}+(-2)^{-4}= - \frac{1}{2} + \frac{1}{2^2} - \frac{1}{2^3} + \frac{1}{2^4} = \\ \\ =- \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} =- \frac{8}{16} + \frac{4}{16} - \frac{2}{16} + \frac{1}{16} =- \frac{5}{16}$

$\displaystyle b).[(-5)^{-1}]^0+5^{-2}-0,2^{-1}=1+ \frac{1}{5^2} -\left( \frac{2}{10} \right)^{-1}=1+ \frac{1}{25} - \frac{10}{2} = \\ \\ =1+ \frac{1}{25} -5= \frac{25+1-125}{25} =- \frac{99}{25}$

$\displaystyle c).4^{-3}+[(-2)^{-1}]^2-2^{-2}= \frac{1}{4^3} + \left(- \frac{1}{2} \right)^2- \frac{1}{2^2} = \frac{1}{64} + \frac{1}{4} - \frac{1}{4} = \frac{1}{64}$

$\displaystyle d).81^{-1}+9^{-2}-[(-3)^{-4}]^0= \frac{1}{81} + \frac{1}{9^2} -1= \frac{1}{81}+ \frac{1}{81} -1= \\ \\ = \frac{1+1-81}{81} =- \frac{79}{81}$

$\displaystyle e).3^2:7+7^{-2} \cdot 5^3-\left(2 \frac{1}{3} \right)^{-2}= 9:7+ \frac{1}{7^2} \cdot 125-\left( \frac{7}{3} \right)^{-2}= \\ \\ =9 \cdot \frac{1}{7} + \frac{1}{49} \cdot 125 - \frac{3^2}{7^2} = \frac{9}{7} + \frac{125}{49} - \frac{9}{49} = \frac{63+125-9}{49} = \frac{179}{49}$

$\displaystyle f).0,25^{-2} \cdot (0,5^{-2}+3 \cdot 0,2^{-2})=\left( \frac{25}{100} \right)^{-2} \cdot \left[\left( \frac{5}{10} \right)^{-2}+3 \cdot \left( \frac{2}{10} \right)^{-2\right] \\ \\ = \frac{100^2}{25^2} \cdot \left( \frac{10^2}{5^2} +3 \cdot \frac{10^2}{2^2} \right)= \frac{10000}{625} \cdot \left( \frac{100}{25} +3 \cdot \frac{100}{4}\right) = \\ \\ = \frac{10000}{625} \cdot \left( \frac{100}{25} + \frac{300}{4} \right)= 16 \cdot (4+75)=16 \cdot 79=1264$

$\displaystyle g).[12^{-2}\cdot (3^2+3)+2^{-3}] \cdot 5^{-1}=\left[ \frac{1}{12^2} \cdot (9+3)+ \frac{1}{2^3} \right]\cdot \frac{1}{5} = \\ \\ =\left( \frac{1}{144} \cdot 12+ \frac{1}{8} \right) \cdot \frac{1}{5} = \left(\frac{12}{144} + \frac{1}{8} \right) \cdot \frac{1}{5} = \left( \frac{12}{144} + \frac{18}{144}\right)\cdot \frac{1}{5} = \\ \\ = \frac{30}{144} \cdot \frac{1}{5} = \frac{30}{720} = \frac{1}{24}$

$\displaystyle h). \frac{(-4) \cdot (-8)^2\cdot 16^3}{\left( \frac{1}{32} \right)^{-10}} :2^{ -5^{2} } +0,25= \frac{-4 \cdot 64 \cdot 4096}{32^1^0} : \frac{1}{2^2^5} + \frac{25}{100} = \\ \\ = \frac{-2^2 \cdot 2^6\cdot 2^1^2}{32^1^0} \cdot 2^2^5+ \frac{25}{100} = \frac{-2^{2+6+12}}{32^1^0} \cdot 2^2^5+ \frac{25}{100} = \\ \\ = \frac{-2^2^0}{32^1^0} \cdot 2^2^5+ \frac{25}{100} =- \frac{2^4^5}{32^1^0} + \frac{25}{100}$