Question

Ajutor dau 50 de puncte

Answer 1

$a)[( 1\frac{1}{2})^{-2} +(2 \frac{1}{4})^{-1} -19( \frac{1}{17})^{0}]:( \frac{1}{3})^{2}= \\ \\ (1\frac{1}{2})^{-2} = (\frac{2+1}{2})^{-2}= (\frac{3}{2}) ^{-2} = (\frac{2}{3}) ^{2}= \frac{4}{9} \\ \\(2 \frac{1}{4})^{-1}= (\frac{2\cdot4+1}{2})^{-1}= (\frac{9}{4}) ^{-1} = \frac{4}{9} \\ \\19( \frac{1}{17})^{0}=1 \\ \\ (\frac{4}{9} +\frac{4}{9} -1): ( \frac{1}{9}) \\ \\ \frac{8-9}{9}\cdot \frac{9}{1}=1$

$b) {[( 2^{-1}+3^{-1}- \frac{1}{4}+ \frac{5}{6})\cdot \frac{6}{17}+ \frac{5}{4}]^{-1}- \frac{3}{14}}:\frac{5}{7} = \\ \\ {[( \frac{1}{2}+ \frac{1}{3} - \frac{1}{4}+ \frac{5}{6})\cdot \frac{6}{17}+ \frac{5}{4}]^{-1}- \frac{3}{14}}\cdot\frac{7}{5} = \\ \\ {[( \frac{6+4-3+10}{12})\cdot \frac{6}{17}+ \frac{5}{4}]^{-1} - \frac{3}{10}= \\ \\ {[( \frac{17}{12})\cdot \frac{6}{17}+ \frac{5}{4}]^{-1} -\frac{3}{10}=\\ \\ (\frac{1}{2} + \frac{5}{4}]^{-1} - \frac{3}{10}$
$(\frac{7}{4})^{-1} - \frac{3}{10}= \frac{40-21}{70} = \frac{19}{70}$

$c)[( \frac{3}{5})^{2}] ^{3} = (\frac{3}{5}})^{6} = \frac{3^{6}}{5^{6}}$
$\[(\frac{3}{5})^{2}]^{3}= \frac{3^{8}}{5^{8}} ; \frac{3^{13}}{5^{13} }$

$\frac{3^{6}\cdot3^{8} }{5^{6}\cdot5^{8} }: \frac{3^{13}}{5^{13}}+ \frac{2}{5}= \frac{ 3^{14}}{5 ^{14}}\cdot \frac{5^{13}}{3^{13}}+ \frac{2}{5} =\\ \\ = \frac{3}{5}+ \frac{2}{5}= \frac{5}{5}=1$

d)
$\frac{3}{8}^{3}= \frac{3^{3}}{2^{9}} \\ \\(\frac{16}{15})^{3}= \frac{(2^{4})^{3}}{3^{3}\cdot5^{3}} = \frac{(2^{12})}{3^{3}\cdot5^{3}} \\ \\ (\frac{35}{10})^{3}= \frac{(7\cdot5)^{3} } {2^{3}\cdot 5^{3}}\\ \\ \frac{3^{3}}{2^{9}} \cdot \frac{(2^{12})}{3^{3}\cdot5^{3}}\cdot \frac{7^{3} \cdot5 ^{3} } {2^{3}\cdot 5^{3}} :[ \frac{14^{2} }{11^{2}} \cdot \frac{11^{2} }{2^{2}\cdot3^{2}}\cdot \frac{3^{2}}{5^{2}} ] - \frac{2}{5}$

$\frac{3^{3}}{2^{9}} \cdot \frac{(2^{12})}{3^{3}\cdot5^{3}}\cdot \frac{7^{3} \cdot5 ^{3} } {2^{3}\cdot 5^{3}} = \frac{ 7^{3}}{5^{3}}$

$[ \frac{14^{2} }{11^{2}} \cdot \frac{11^{2} }{2^{2}\cdot3^{2}}\cdot \frac{3^{2}}{5^{2}} ]= \frac{ 7^{2}}{5^{2}}$

$=\frac{ 7^{3}}{5^{3}}:\frac{ 7^{2}}{5^{2}}- \frac{2}{5} \\ \\ \frac{ 7^{3}}{5^{3}}\cdot\frac{ 5^{2}}{7^{2}}- \frac{2}{5} = \\ \\ = \frac{7}{5} - \frac{2}{5} = \frac{5}{5}=1$