Matemática, perguntado por Manu1882, 9 meses atrás

efetue as adiçoes:
C)1/6+3/6-7/6=
D)3/4+1/4+7/4=
E)1/9-3/9-5/9=
F)4/3-1/3-2/3=
G)8/5-10/5-1/5=
H)1/7+2/7-17/7=
I)-3/5+2/5+8/5=
J)-2/6-1/6+3/6=
ALGUEM ME AJUDA PFV??

Soluções para a tarefa

Respondido por netinbookoxmu3a
2

Resposta:

Explicação passo-a-passo:

\dfrac{1}{6}+\dfrac{3}{6}-\dfrac{7}{6}=\dfrac{1+3-7}{6}=\dfrac{-3}{6}=-\dfrac{1}{2}

\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{3+1}{4}=\dfrac{4}{4}=1

\dfrac{1}{9}-\dfrac{3}{9}-\dfrac{5}{9}=\dfrac{1-3-5}{9}=\dfrac{-7}{9}

\dfrac{4}{3}-\dfrac{1}{3}-\dfrac{2}{3}=\dfrac{4-1-2}{3}=\dfrac{1}{3}

\dfrac{8}{5}-\dfrac{10}{5}-\dfrac{1}{5}=\dfrac{8-10-1}{5}=\dfrac{-3}{5}

\dfrac{1}{7}+\dfrac{2}{7}-\dfrac{17}{7}=\dfrac{1+2-17}{7}=\dfrac{-14}{7}=-2

\dfrac{-3}{5}+\dfrac{2}{5}+\dfrac{8}{5}=\dfrac{-3+2+8}{5}=\dfrac{7}{5}

\dfrac{-2}{6}-\dfrac{1}{6}+\dfrac{3}{6}=\dfrac{-2-1+3}{6}=\dfrac{0}{6}=0

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