Question

Ma puteti ajuta va rog cu tot ce este in poza si ce am scris mai jos
C)11/9-[3/4+(- 5/36)]
D) (- 3/14)+[(- 1/2)+(+3/7)+(-5/14)]

Answer 1

Răspuns:

Explicație pas cu pas:

e.

$-3\frac{1}{5} +(-2\frac{1}{6} +\frac{3}{2} )=$

$=-\frac{16}{5} +(-\frac{13}{6} +\frac{9}{6} )=$

$=-\frac{16}{5} +\frac{-13+9}{6}=$

$=-\frac{16}{5} -\frac{4}{6}=$

$=-\frac{96}{30} -\frac{20}{30}=$

$=-\frac{116}{30} =$

$=-3\frac{26}{30} =$

$=-3\frac{13}{15}$

f.

$\frac{8}{15} -(-1\frac{2}{3} +2\frac{1}{5} ) + \frac{7}{15} =$

$=\frac{8}{15} -(-\frac{5}{3} +\frac{11}{5} ) + \frac{7}{15} =$

$=\frac{8}{15} -(-\frac{25}{15} +\frac{33}{15} ) + \frac{7}{15} =$

$=\frac{8}{15} -\frac{8}{15} + \frac{7}{15} =$

$=\frac{7}{15}$

3.

a.

$\frac{1}{3} +\frac{3}{4} +(\frac{2}{3} +\frac{5}{6} ) =$  

$= \frac{4}{12} +\frac{9}{12} +(\frac{4}{6} +\frac{5}{6} ) =$

$= \frac{13}{12}+\frac{9}{6}=$

$= \frac{13}{12}+\frac{18}{12}=$

$= \frac{31}{12}$

b.

$\frac{7}{15} -(\frac{3}{5} +\frac{4}{3} +\frac{1}{15} ) =$

$=\frac{7}{15} -(\frac{9}{15} +\frac{20}{15} +\frac{1}{15} ) =$

$=\frac{7}{15} -\frac{30}{15} =$

$=-\frac{23}{15}$

c.

$\frac{11}{9} -[\frac{3}{4} +(-\frac{5}{36} )]=$

$=\frac{11}{9} -(\frac{3}{4} -\frac{5}{36} )=$

$=\frac{11}{9} -(\frac{27}{36} -\frac{5}{36} )=$

$=\frac{11}{9} -\frac{22}{36}=$

$=\frac{11}{9} -\frac{11}{18}=$

$=\frac{22}{18} -\frac{11}{18}=$

$=\frac{11}{18}$

d.

$(-\frac{3}{14})+[(-\frac{1}{2}) +(+\frac{3}{7}) +(-\frac{5}{14} )] =$

$-\frac{3}{14}+(-\frac{1}{2} +\frac{3}{7} -\frac{5}{14}) =$

$-\frac{3}{14}+(-\frac{7}{14} +\frac{6}{14} -\frac{5}{14}) =$

$-\frac{3}{14}+(-\frac{6}{14}) =$

$-\frac{3}{14}-\frac{6}{14} =$

$-\frac{9}{14}$

Answer 2

$\it b)\ \dfrac{7}{15}-\Big(\dfrac{3}{5}+\dfrac{4}{3}+\dfrac{1}{15}\Big)=\dfrac{7}{15}-\dfrac{^{3)}3}{\ 5}-\dfrac{^{5)}4}{\ 3}-\dfrac{1}{15}=\dfrac{7-9-20-1}{15}=-\dfrac{23}{15}$

$\it c)\ \dfrac{11}{9}-\Big[\dfrac{3}{4}+\Big(-\dfrac{5}{36}\Big)\Big]=\dfrac{^{4)}11}{\ 9}-\dfrac{^{9)}3}{\ 4}+\dfrac{5}{36}=\dfrac{44-27+5}{36}=\dfrac{\ 22^{(2}}{36}=\dfrac{11}{18}$

$\it d)\ -\dfrac{3}{14}+\Big[\Big(-\dfrac{1}{2}\Big)+\Big(+\dfrac{3}{7}\Big)+\Big(-\dfrac{5}{14}\Big)\Big]=-\dfrac{3}{14}-\dfrac{^{7)}1}{\ 2}+\dfrac{^{2)}3}{\ 7}-\dfrac{5}{14}=\\ \\ \\ =\dfrac{-3-7+6-5}{14}=-\dfrac{9}{14}$