Question

38. Completați spațiile punctate: a) 0,(3) - 1 ( (0:0) b) + 3 39. Efectuând calculele, se obține: 3 3 (-1)2019 a) (-1) 4 5 2 -1 + 1 b) Dacă n e N, atunci 3 - (+1) ++ (-1) 40. Fie S, suma elementelor mulțimii A = {n A și S, suma elementelor mulțimii B = {n\ n = 4k, B. Soluţie: ....​

Answer 1

a)

$[ 0.(3) - 1 \frac{1}{2} ] \times {( - \frac{1}{3}) }^{ - 1} = \\ ( \frac{3}{9} - \frac{1 \times 2 + 1}{2} ) \times ( - \frac{3}{1} ) = \\ ( \frac{1}{3} - \frac{3}{2} ) \times ( - 3) = \\ ( \frac{2}{6} - \frac{9}{6} ) \times ( - 3) = \\( - \frac{7}{6} ) \times (- \frac{3}{1} ) = \\ ( - \frac{7}{2} ) \times ( - 1) = \frac{7}{2}$

b)

${( - \frac{1}{2} )}^{4} \times {( - \frac{1}{2} )}^{3} \div {( - \frac{1}{2} )}^{6} + {( \frac{2}{3}) }^{ - 1} =$

${( - \frac{1}{2} )}^{4 + 3} \div {( - \frac{1}{2} )}^{6} + {( \frac{3}{2}) }^{ }$

${( - \frac{1}{2} )}^{7} \div {( - \frac{1}{2} )}^{6} + { \frac{3}{2} }^{ }$

${( - \frac{1}{2} )}^{7 - 6} + { \frac{3}{2} }^{ }$

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

Answer 2

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