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\(a=\frac12+\frac56+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}\)
\(=1-\frac12+1-\frac16+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}\)
\(=7-\left(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(=7-\left(1-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\frac17-\frac18\right)\)
\(=7-\left(1-\frac18\right)=7-\frac78=\frac{56}{8}-\frac78=\frac{49}{8}\)
Đề bài:
a = 1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56
Giải:
Ta thấy mỗi phân số có tử số = mẫu số trừ 1
Vậy ta viết lại:
a = (2 - 1)/2 + (6 - 1)/6 + (12 - 1)/12 + (20 - 1)/20 + (30 - 1)/30 + (42 - 1)/42 + (56 - 1)/56
= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 + 1 - 1/20 + 1 - 1/30 + 1 - 1/42 + 1 - 1/56
= 7 - (1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56)
= 7 - 3/2
= 11/2
Đáp số: 11/2
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\(=1-\frac{1}{1\cdot2}+1-\frac{1}{2\cdot3}+1-\frac{1}{3\cdot4}+...+1-\frac{1}{9\cdot10}\)
\(=\left[1+1+1+...+1\right]-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right]\)
\(=9-\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right]=9-\left[1-\frac{1}{10}\right]\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{!}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10
= 81/10
\(y=\frac{1+5+11+19+29+41+55+71+89}{2+6+12+20+30+42+56+72+90}\)
\(y=\frac{1x2-1+2x3-1+3x4-1+4x5-1+5x6-1+6x7-1+7x8-1+8x9-1+9x10-1}{1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9+9x10}\)
\(y=\frac{\left(1x2+2x3+...+9x10\right)-\left(1+1+1+1+1+1+1+1+1\right)}{1x2+2x3+...+9x10}\)
\(y=\frac{1x2+2x3+...+9x10}{1x2+2x3+...+9x10}-\frac{9}{1x2+2x3+...+9x10}\)
\(y=1-\frac{9}{1x2+2x3+...+9x10}\)
\(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)
\(=\frac{12-1}{12}+\frac{20-1}{20}+\frac{30-1}{30}+\frac{42-1}{42}+\frac{56-1}{56}+\frac{72-1}{72}+\frac{90-1}{90}+\frac{110-1}{110}\)
\(=1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=8-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{256}{33}\)
ĐẶT A = \(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{109}{110}\)
8 -A = \(1-\frac{11}{12}+1-\frac{19}{20}+1-\frac{29}{30}+1-\frac{41}{42}+1-\frac{55}{56}+1-\frac{71}{72}+1-\frac{89}{90}+1-\frac{109}{110}\)
8 -A \(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
8 -A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
8 - A = \(\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
=> A = \(8-\frac{8}{33}=\frac{256}{33}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)
\(=(1-\frac{1}{2})+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\) \(+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\frac{11}{10}\)
\(=\frac{79}{10}\)
~Học tốt nha~
Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(\Leftrightarrow A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+......+\left(1-\frac{1}{90}\right)\)
\(\Leftrightarrow A=\left(1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{90}\right)\)
\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(\Leftrightarrow A=9-\left(1-\frac{1}{10}\right)\)
\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)
bn vào câu hỏi tương tự sẽ có chi tiết . Nếu k thì bn hãy để ý mỗi tử đều bé hơn mẫu 1 đơn vị sau đó bn tách ra bằng cách lấy 1 trừ . VD: 5/6 bằng 1 - 1/6 . Đến đó đếm đc 9 chữ số 1 ta lấy 9 làm sbt trừ đi tổng của các ps ta tách đc . Khi đó thì bài toán quá đơn giản rồi . Chúc bn học tốt
(1-1/2)+(1-1/6)+...+(1-1/90)
9+(1/2+1/6+...+1/90)
9+(1/1.2+1/2.3+...+1/9.10)
9+1-9/10=9/1/10=91/10
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