N° 3

Question

-2/3-{1/6-[(1/2-1/3)+3-(7/12+1/4)]}

gramor · Accepted Answer

-2/3-{1/6-[(1/2-1/3)+3-(7/12+1/4)]}

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$-\frac{2}{3}-\big\{\frac{1}{6}-\big[\big(\frac{1}{2}-\frac{1}{3}\big)+3-\big(\frac{7}{12}+\frac{1}{4}\big)\big]\big\}=$

$= -\frac{2}{3}-\big\{\frac{1}{6}-\big[\big(\frac{3-2}{6}\big)+3-\big(\frac{7+3}{12}\big)\big]\big\}=$

$= -\frac{2}{3}-\big\{\frac{1}{6}-\big[\big(\frac{1}{6}\big)+3-\big(\frac{10}{12}\big)\big]\big\}=$

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

$= -\frac{2}{3}-\big\{\frac{1}{6}-\big[\frac{1+18-5}{6}\big]\big\}=$

$= -\frac{2}{3}-\big\{\frac{1}{6}-\frac{14}{6}\big\}=$

$= -\frac{2}{3}-\big\{-\frac{13}{6}\big\}=$

$= -\frac{2}{3}+\frac{13}{6}=$

$= \frac{-4+13}{6}=$

$= \frac{9}{6}=$

$= \frac{3}{2}$

gramor

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18/11/2023 20:26

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