Question

Vă rog să mă ajutați cu ex. 7, 8 îmi pune notă, dau coronita.

Answer 1

$a)\\a=(-3\frac{3}{8})-(-2\frac{1}{6})-(1\frac{3}{4})=(\frac{-3\cdot8+3}{8})-(\frac{2\cdot6+1}{6}-(\frac{1\cdot4+3}{4})=\frac{-27}{8}+\frac{13}{6}-\frac{7}{4}=\frac{-27\cdot3+4\cdot13-42}{24}=-\frac{71}{24}\\ b=(2\frac{5}{6})-(-1\frac{2}{3})-(3\frac{1}{4})=(\frac{2\cdot6+5}{6})-(-\frac{1\cdot3+2}{3}-(-\frac{1\cdot3+2}{4})=\frac{-17}{6}+\frac{5}{3}-\frac{13}{4}=\frac{4\cdot17+8\cdot5-13}{4}=-\frac{68+40-78}{24}=\frac{30}{24}$

$a-b=-\frac{71}{24}-\frac{30}{24}=-\frac{101}{24}$

$\\b)a=(-1\frac{3}{14})-(2\frac{4}{21})-(4\frac{5}{6})=(-\frac{1\cdot14+3}{14})-(\frac{2\cdot21+4}{21}-(-\frac{4\cdot6+5}{6})=\frac{-17}{14}-\frac{46}{21}+\frac{29}{6}=\frac{-17\cdot6-4\cdot46+29\cdot14}{12\cdot7}=-\frac{-102-184+406}{84}=\frac{120}{84}=\frac{10}{7}\\ b=(-4\frac{1}{2})-(3\frac{1}{6})-(-5\frac{2}{3})=(-\frac{4\cdot2+1}{2})-(\frac{3\cdot6+1}{6}+-\frac{5\cdot3+2}{3})=\frac{-9}{2}-\frac{19}{6}+\frac{17}{3}=\frac{-9\cdot3-19+17\cdot2}{6}=\frac{-12}{6}=-2$

$a-b=\frac{10}{7}-(-2)=\frac{10}{7}+2=\frac{24}{7}=3\frac{3}{7}$

$8.a)(-\frac{1}{10}-(-\frac{2}{15})-[(-\frac{1}{30})+(+\frac{5}{6})]}=(-\frac{1}{10}+\frac{2}{15})-[-\frac{1}{30}+\frac{5}{6}]}=\frac{-1\cdot3+2\cdot2}{30}-(\frac{-1+5\cdot5}{30}=\frac{1}{30}+\frac{9}{30}=\frac{10}{30}=\frac{1}{3}$

$b)(-\frac{3}{28}-[(-\frac{5}{42})-(+\frac{1}{14})+(-\frac{2}{21})]}=(-\frac{3}{28})-[(-\frac{5}{42})-[\frac{1}{14}-\frac{2}{21}]}=(-\frac{3}{28})-[(-\frac{5}{42})-[\frac{1\cdot3-2\cdot2}{2\cdot3\cdot7}}]}=(-\frac{3}{28})-[(-\frac{5}{42})-\frac{1}{42}}}=(-\frac{3}{28})-(-\frac{6}{42})}=(-\frac{3}{28})-(-\frac{1}{7})}=\frac{-3+1\cdot4}{28}=\frac{1}{28}$