dạng toán mà dãy số có dạng a/a+1 A=5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90 09/08/2021 Bởi Arianna dạng toán mà dãy số có dạng a/a+1 A=5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
Đáp án: Ta có : A = $\dfrac{5}{6}$ + $\dfrac{11}{12}$ + $\dfrac{19}{20}$ + $\dfrac{29}{30}$ + $\dfrac{41}{42}$ + $\dfrac{55}{56}$ + $\dfrac{71}{72}$ + $\dfrac{89}{90}$ => A = 1 – $\dfrac{1}{6}$ + 1 – $\dfrac{1}{12}$ + 1 – $\dfrac{1}{20}$ + 1 – $\dfrac{1}{30}$ + 1 – $\dfrac{1}{42}$ + 1 – $\dfrac{1}{56}$ + 1 – $\dfrac{1}{72}$ + 1 – $\dfrac{1}{90}$ => A = 8 – ( $\dfrac{1}{6}$ + $\dfrac{1}{12}$ + $\dfrac{1}{20}$ + $\dfrac{1}{30}$ + $\dfrac{1}{42}$ + $\dfrac{1}{56}$ + $\dfrac{1}{72}$ + $\dfrac{1}{90}$ Đặt B = $\dfrac{1}{6}$ + $\dfrac{1}{12}$ + $\dfrac{1}{20}$ + $\dfrac{1}{30}$ + $\dfrac{1}{42}$ + $\dfrac{1}{56}$ + $\dfrac{1}{72}$ + $\dfrac{1}{90}$ => B = $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + $\dfrac{1}{4.5}$ + $\dfrac{1}{5.6}$ + $\dfrac{1}{6.7}$ + $\dfrac{1}{7.8}$ + $\dfrac{1}{8.9}$ + $\dfrac{1}{9.10}$ = $\dfrac{1}{2}$ – $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{4}$ + …. + $\dfrac{1}{9}$ – $\dfrac{1}{10}$ = $\dfrac{1}{2}$ – $\dfrac{1}{10}$ = $\dfrac{2}{5}$ => A = 8 – $\dfrac{2}{5}$ = $\dfrac{19}{20}$ Giải thích các bước giải: Bình luận
A = 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90 A = (1 – 1/6) + (1 – 1/12) + … + ( 1 – 1/90) A = (1 . 8) – (1/6 + 1/12 + 1/20 + … + 1/90) A = 8 – (1/2.3 + 1/3.4 + 1/4.5 + … + 1/9.10) A = 8 – (1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/9 – 1/10) A = 8 – (1/2 – 1/10) A = 8 – 2/5 A = 38/5 Bình luận
Đáp án:
Ta có :
A = $\dfrac{5}{6}$ + $\dfrac{11}{12}$ + $\dfrac{19}{20}$ + $\dfrac{29}{30}$ + $\dfrac{41}{42}$ + $\dfrac{55}{56}$ + $\dfrac{71}{72}$ + $\dfrac{89}{90}$
=> A = 1 – $\dfrac{1}{6}$ + 1 – $\dfrac{1}{12}$ + 1 – $\dfrac{1}{20}$ + 1 – $\dfrac{1}{30}$ + 1 – $\dfrac{1}{42}$ + 1 – $\dfrac{1}{56}$ + 1 – $\dfrac{1}{72}$ + 1 – $\dfrac{1}{90}$
=> A = 8 – ( $\dfrac{1}{6}$ + $\dfrac{1}{12}$ + $\dfrac{1}{20}$ + $\dfrac{1}{30}$ + $\dfrac{1}{42}$ + $\dfrac{1}{56}$ + $\dfrac{1}{72}$ + $\dfrac{1}{90}$
Đặt B = $\dfrac{1}{6}$ + $\dfrac{1}{12}$ + $\dfrac{1}{20}$ + $\dfrac{1}{30}$ + $\dfrac{1}{42}$ + $\dfrac{1}{56}$ + $\dfrac{1}{72}$ + $\dfrac{1}{90}$
=> B = $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + $\dfrac{1}{4.5}$ + $\dfrac{1}{5.6}$ + $\dfrac{1}{6.7}$ + $\dfrac{1}{7.8}$ + $\dfrac{1}{8.9}$ + $\dfrac{1}{9.10}$
= $\dfrac{1}{2}$ – $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{4}$ + …. + $\dfrac{1}{9}$ – $\dfrac{1}{10}$
= $\dfrac{1}{2}$ – $\dfrac{1}{10}$ = $\dfrac{2}{5}$
=> A = 8 – $\dfrac{2}{5}$ = $\dfrac{19}{20}$
Giải thích các bước giải:
A = 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90
A = (1 – 1/6) + (1 – 1/12) + … + ( 1 – 1/90)
A = (1 . 8) – (1/6 + 1/12 + 1/20 + … + 1/90)
A = 8 – (1/2.3 + 1/3.4 + 1/4.5 + … + 1/9.10)
A = 8 – (1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/9 – 1/10)
A = 8 – (1/2 – 1/10)
A = 8 – 2/5
A = 38/5