Question

VA ROG AJUTATTI-MA!!
EX 3,4
99 DE PUNCTE+COROANA

Answer 1

3.
a. -2+16=14
b. 9-6=3
c. 25+10=35
d. 1/9×(-1/8)=-1/18
e. 0,04+(-0,4)=-0,36
f. -27/8-1/9 . amplicam prima cu 9 si a doua cu8=-244/72-8/72=236/72
g.(-243/1024):(-2187/128)=1/72

Answer 2

$\displaystyle 3a).2^{-1}+4^{-2}= \frac{1}{2} + \frac{1}{4^2} = \frac{1}{2} + \frac{1}{16} = \frac{8}{16} + \frac{1}{16} = \frac{9}{16} \\ \\ b).3^{-2}-6^{-1}= \frac{1}{3^2} - \frac{1}{6} = \frac{1}{9} - \frac{1}{6} = \frac{2}{18} - \frac{3}{18} =- \frac{1}{18}$

$\displaystyle c).5^{-2}-(-10)^{-1}= \frac{1}{5^2} - \left (- \frac{1}{10} \right)= \frac{1}{25} + \frac{1}{10} = \frac{2}{50} + \frac{5}{50} = \frac{7}{50}$

$\displaystyle d).\left( \frac{2}{3} \right)^{-2} \cdot \left( \frac{1}{2} \right)^{-3}= \frac{3^2}{2^2} \cdot 2^3= \frac{9}{4} \cdot 8= \frac{72}{4} =18$

$\displaystyle e).(0,2)^{-2}+(0,4)^{-1}=\left( \frac{2}{10} \right)^{-2}+ \left( \frac{4}{10} \right )^{-1}= \frac{10^ 2}{2^2} + \frac{10}{4} = \frac{100}{4} + \frac{10}{4} = \\ \\ = \frac{110}{4} = \frac{55}{2}$

$\displaystyle f).\left(1 \frac{1}{2} \right )^{-3}-[0,(3)]^2=\left( \frac{3}{2} \right)^{-3}-\left( \frac{3}{9} \right)^2= \frac{2^3}{3^3} - \frac{3^2}{9^2} = \frac{8}{27} - \frac{9}{81} = \\ \\ = \frac{24}{81} - \frac{9}{81} = \frac{15}{81} = \frac{5}{27}$

$\displaystyle g).\left(- \frac{3}{4} \right )^{-5}:\left( \frac{3}{2} \right )^{-7}=- \frac{4^5}{3^5} : \frac{2^7}{3^7} = -\frac{1024}{243} : \frac{128}{2187} =- \frac{1024}{243} \cdot \frac{2187}{128} = \\ \\ = - \frac{2239488}{31104} =-72$

$\displaystyle h).\left(-1 \frac{1}{2} \right)^{-4} \cdot [-0,(3)]^{-3}=\left(- \frac{3}{2} \right)^{-4} \cdot \left(- \frac{3}{9} \right)^{-3}=- \frac{2^4}{3^4} \cdot \left(- \frac{9^3}{3^3} \right)= \\ \\ =- \frac{16}{81} \cdot \left(- \frac{729}{27} \right)= \frac{11664}{2187} = \frac{16}{3}$

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

$\displaystyle b).(-3)^{-2}+[(+3)^{-1}]^2= \frac{1}{3^2} +\left( \frac{1}{3} \right)^2= \frac{1}{9} + \frac{1^2}{3^2} = \frac{1}{9} + \frac{1}{9}= \frac{2}{9}$

$\displaystyle c).0,2^{-3}-\left( \frac{1}{11} \right)^{-2}=\left( \frac{2}{10} \right)^{-3}-11^2= \frac{10^3}{2^3} -121= \frac{1000}{8} -121= \\ \\ =125-121=4$

$\displaystyle d). \frac{5}{6} :6^{-2}-1:5^{-2}= \frac{5}{6} : \frac{1}{6^2} -1: \frac{1}{5^2} = \frac{5}{6} : \frac{1}{36} -1: \frac{1}{25} = \\ \\ = \frac{5}{6} \cdot 36-1 \cdot 25= \frac{180}{6} -25=30-25=5$

$\displaystyle e).(2^{-3}+2^{-4}) \cdot 3^{-1}=\left( \frac{1}{2^3} + \frac{1}{2^4} \right) \cdot \frac{1}{3} =\left( \frac{1}{8} + \frac{1}{16} \right)\cdot \frac{1}{3} = \\ \\ =\left( \frac{2}{16} + \frac{1}{16} \right)\cdot \frac{1}{3} = \frac{3}{16} \cdot \frac{1}{3} = \frac{3}{48} = \frac{1}{16}$

$\displaystyle f).12^{-2} \cdot (-6)^2-2^2 \cdot 4^{-2}= \frac{1}{12^2} \cdot 36-4 \cdot \frac{1}{4^2} = \frac{1}{144} \cdot 36-4 \cdot \frac{1}{16} = \\ \\ = \frac{36}{144} - \frac{4}{16} = \frac{36}{144} - \frac{36}{144} =0$