Question

Đã trả lời bởi chuyên gia

Answer 1

Ta tính lần lượt từng biểu thức.

A.

$A = (3 - \frac{1}{4} + \frac{2}{3}) - (5 \frac{1}{3} - \frac{6}{5}) - (6 - \frac{7}{4} + \frac{3}{2}) Đ ổ i s ố h ỗ n h ợ p : 5 \frac{1}{3} = \frac{16}{3} T í n h t ừ n g n g o ặ c N g o ặ c 1 3 - \frac{1}{4} + \frac{2}{3} = \frac{36 - 3 + 8}{12} = \frac{41}{12} . N g o ặ c 2 \frac{16}{3} - \frac{6}{5} = \frac{80 - 18}{15} = \frac{62}{15} N g o ặ c 3 6 - \frac{7}{4} + \frac{3}{2} = \frac{24 - 7 + 6}{4} = \frac{23}{4} D o đ ó A = \frac{41}{12} - \frac{62}{15} - \frac{23}{4} Q u y đ ồ n g m ẫ u s ố 60 : A = \frac{205 - 248 - 345}{60} = \frac{- 388}{60} = - \frac{97}{15} Đ á p s ố A = - \frac{97}{15}$

B.

$B = \frac{1}{3} - \frac{3}{4} - (- \frac{3}{5}) + \frac{1}{64} - \frac{2}{9} - \frac{1}{36} + \frac{1}{15} T a c ó - (- \frac{3}{5}) = + \frac{3}{5} N h ó m c á c p h â n s ố : \frac{1}{3} - \frac{2}{9} = \frac{1}{9} . \frac{1}{9} - \frac{1}{36} = \frac{4}{36} - \frac{1}{36} = \frac{1}{12} . \frac{1}{12} + \frac{1}{15} = \frac{15}{20} = \frac{3}{4} . V ậ y B = \frac{3}{4} - \frac{3}{4} + \frac{1}{64} = \frac{1}{64}$

C.

$C = \frac{1}{3} - \frac{3}{5} + \frac{5}{7} - \frac{7}{9} + \frac{9}{11} - \frac{11}{13} + \frac{13}{15} + \frac{11}{13} - \frac{9}{11} + \frac{7}{9} . C á c c ặ p t r i ệ t t i ê u - \frac{7}{9} + \frac{7}{9} = 0 \frac{9}{11} - \frac{9}{11} = 0, - \frac{11}{13} + - \frac{11}{13} = 0 C ò n l ạ i C = \frac{1}{3} - \frac{3}{5} + \frac{5}{7} + \frac{13}{15} . T a c ó \frac{1}{3} + \frac{13}{15} = \frac{5}{15} + \frac{13}{15} = \frac{18}{15} = \frac{6}{5} N ê n C = \frac{6}{5} - \frac{3}{5} + \frac{5}{7} = \frac{3}{5} + \frac{5}{7} = \frac{21 + 25}{35} = \frac{46}{35}$

D.

$D = \frac{1}{99} - \frac{1}{99.98} - \frac{1}{99.97} - . . . - \frac{1}{3.2} - \frac{1}{2.1} . T a s ử d ụ n g \frac{1}{n (n - 1)} = \frac{1}{n - 1} - \frac{1}{n} . D o đ ó \sum_{n = 2}^{99} \frac{1}{n (n - 1)} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . . + (\frac{1}{98} - \frac{1}{99}) = 1 - \frac{1}{99} = \frac{98}{99} S u y r a D = \frac{1}{99} - \frac{98}{99} = - \frac{97}{99} . Đ á p s ố D = - \frac{97}{99} K ế t q u ả c u ố i c ù n g A = - \frac{97}{15} B = \frac{1}{64} C = \frac{46}{35} D = - \frac{97}{99}$

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Answer 2

Ta tính lần lượt từng biểu thức.

A.

$A = (3 - \frac{1}{4} + \frac{2}{3}) - (5 \frac{1}{3} - \frac{6}{5}) - (6 - \frac{7}{4} + \frac{3}{2}) Đ ổ i s ố h ỗ n h ợ p : 5 \frac{1}{3} = \frac{16}{3} T í n h t ừ n g n g o ặ c N g o ặ c 1 3 - \frac{1}{4} + \frac{2}{3} = \frac{36 - 3 + 8}{12} = \frac{41}{12} . N g o ặ c 2 \frac{16}{3} - \frac{6}{5} = \frac{80 - 18}{15} = \frac{62}{15} N g o ặ c 3 6 - \frac{7}{4} + \frac{3}{2} = \frac{24 - 7 + 6}{4} = \frac{23}{4} D o đ ó A = \frac{41}{12} - \frac{62}{15} - \frac{23}{4} Q u y đ ồ n g m ẫ u s ố 60 : A = \frac{205 - 248 - 345}{60} = \frac{- 388}{60} = - \frac{97}{15} Đ á p s ố A = - \frac{97}{15}$

B.

$B = \frac{1}{3} - \frac{3}{4} - (- \frac{3}{5}) + \frac{1}{64} - \frac{2}{9} - \frac{1}{36} + \frac{1}{15} T a c ó - (- \frac{3}{5}) = + \frac{3}{5} N h ó m c á c p h â n s ố : \frac{1}{3} - \frac{2}{9} = \frac{1}{9} . \frac{1}{9} - \frac{1}{36} = \frac{4}{36} - \frac{1}{36} = \frac{1}{12} . \frac{1}{12} + \frac{1}{15} = \frac{15}{20} = \frac{3}{4} . V ậ y B = \frac{3}{4} - \frac{3}{4} + \frac{1}{64} = \frac{1}{64}$

C.

$C = \frac{1}{3} - \frac{3}{5} + \frac{5}{7} - \frac{7}{9} + \frac{9}{11} - \frac{11}{13} + \frac{13}{15} + \frac{11}{13} - \frac{9}{11} + \frac{7}{9} . C á c c ặ p t r i ệ t t i ê u - \frac{7}{9} + \frac{7}{9} = 0 \frac{9}{11} - \frac{9}{11} = 0, - \frac{11}{13} + - \frac{11}{13} = 0 C ò n l ạ i C = \frac{1}{3} - \frac{3}{5} + \frac{5}{7} + \frac{13}{15} . T a c ó \frac{1}{3} + \frac{13}{15} = \frac{5}{15} + \frac{13}{15} = \frac{18}{15} = \frac{6}{5} N ê n C = \frac{6}{5} - \frac{3}{5} + \frac{5}{7} = \frac{3}{5} + \frac{5}{7} = \frac{21 + 25}{35} = \frac{46}{35}$

D.

$D = \frac{1}{99} - \frac{1}{99.98} - \frac{1}{99.97} - . . . - \frac{1}{3.2} - \frac{1}{2.1} . T a s ử d ụ n g \frac{1}{n (n - 1)} = \frac{1}{n - 1} - \frac{1}{n} . D o đ ó \sum_{n = 2}^{99} \frac{1}{n (n - 1)} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . . + (\frac{1}{98} - \frac{1}{99}) = 1 - \frac{1}{99} = \frac{98}{99} S u y r a D = \frac{1}{99} - \frac{98}{99} = - \frac{97}{99} . Đ á p s ố D = - \frac{97}{99} K ế t q u ả c u ố i c ù n g A = - \frac{97}{15} B = \frac{1}{64} C = \frac{46}{35} D = - \frac{97}{99}$

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