Question

2 supra 3+(-1 sura 4);
-5supra12+-3pe 16
-3pe10+-4pe15
9pe20+-8pe16

Answer 1

$\frac{2}{3} + ( \frac{-1}{4}) = \frac{2*4}{3*4}+( \frac{-1*3}{4*3})= \frac{8}{12}- \frac{3}{12} = \frac{5}{12}$
$\frac{-5}{12} + \frac{3}{16} = \frac{-5*4}{12*4}+ \frac{3*3}{16*3}= \frac{-20}{48}+ \frac{9}{48}= \frac{-11}{48}$
$\frac{-3}{10} + \frac{-4}{15} = \frac{-3*3}{10*3} + \frac{-4*2}{15*2} = (\frac{-9}{30})+ (\frac{-8}{30} )= \frac{-17}{30}$
[tex] \frac{9}{20}+( \frac{-8}{16})= \frac{9}{20} +( \frac{-1}{2} )= \frac{9}{20}+( \frac{-1*10}{2*10})= \frac{9}{20}+( \frac{-10}{20})= \frac{9-10}{20}= \frac{-1}{20} [/tex]$(\frac{-8}{16} simplificam prin 8 si primim \frac{-1}{2}$)