Zadanie 3 podpunkt d) e) f)Zadanie 4 podpunkt a)Zadanie 4 podpunkt d) e) f) Mają być rozwiązania !! .... Question from @matt1423

November 2018 1 23 Report

Zadanie 3 podpunkt d) e) f)
Zadanie 4 podpunkt a)
Zadanie 4 podpunkt d) e) f)

Mają być rozwiązania !! NIE KOPIOWAC Z NETA :)
Dam naaj :)
Już jest załącznik.

Zgłoś nadużycie! 3.
d)
$\frac{7}{6}+\left(- \frac{5}{4}- \frac{4}{3} \right)=\frac{14}{12}+\left(- \frac{15}{12}- \frac{16}{12} \right)=$

$\frac{14}{12}+\left(- \frac{31}{12} \right)=\frac{14}{12}- \frac{31}{12}=- \frac{17}{12}=-1 \frac{5}{12}$
------------------
e)
$-\left(-2 \frac{2}{3} \right)+ \frac{7}{15}=2 \frac{10}{15}+ \frac{7}{15}=2 \frac{17}{15}=3 \frac{2}{15}$
------------------
f)
$-1 \frac{9}{16}+ \frac{11}{12}=-1 \frac{27}{48}+ \frac{44}{48}=-\frac{75}{48}+ \frac{44}{48}=- \frac{31}{48}$
==================
4.
a)
$-2 \frac{1}{3}\cdot \left(- \frac{27}{28} \right)=-\frac{7}{3}\cdot \left(- \frac{27}{28} \right)= \frac{9}{4}=2 \frac{1}{4}$
==================
5.
d)
$\frac{2- \frac{1}{3} }{2- \frac{1}{2} } =\frac{1\frac{2}{3} }{1 \frac{1}{2} } = \frac{\frac{5}{3} }{\frac{3}{2} }= \frac{5}{3}:\frac{3}{2}=\frac{5}{3}\cdot \frac{2}{3}= \frac{10}{9}=1 \frac{1}{9}$
------------------
e)
$\frac{\frac{3}{4}+ \frac{1}{8} }{\frac{3}{4}- \frac{1}{8} }=\frac{\frac{6}{8}+ \frac{1}{8} }{\frac{6}{8}- \frac{1}{8} }=\frac{\frac{7}{8}}{\frac{5}{8}}= \frac{7}{8}:\frac{5}{8}=\frac{7}{8}\cdot \frac{8}{5}= \frac{7}{5}=1 \frac{2}{5}$
------------------
f)
$\frac{\frac{1}{2}-1 \frac{2}{5} }{\frac{1}{2}-1 \frac{1}{5} }=\frac{\frac{5}{10}-1 \frac{4}{10} }{\frac{5}{10}-1 \frac{2}{10} }=\frac{\frac{5}{10}-\frac{14}{10} }{\frac{5}{10}-\frac{12}{10} }=\frac{-\frac{9}{10} }{-\frac{7}{10} }=$

$-\frac{9}{10}:\left(-\frac{7}{10} \right)=-\frac{9}{10}\cdot \left(-\frac{10}{7} \right)= \frac{9}{7}=1 \frac{2}{7}$

1 votes Thanks 1