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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.
k mình nha
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
\(=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{9.10}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
có 9 số 1
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}\)
\(=\frac{81}{10}\)
Ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=8-1+\dfrac{1}{10}\)
\(=\dfrac{71}{10}\)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
Câu hỏi của Nguyễn Ngọc Mai Anh - Toán lớp 5 - Học toán với OnlineMath
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\)\(\frac{55}{66}\)\(+\frac{71}{72}\)\(+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=8+\frac{1}{10}=\frac{81}{10}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
9 - A = 1 - 1/2 + 1-5/6 + 1 - 11/12 + ... + 1-89/90
9 - A = 1/2 + 1/6 + 1/12 + .. + 1/90
9 - A = 1/1.2 + 1/2.3 + 1/3.4 + .. + 1/9.10
9-A = 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10
9 -A = 1/1 - 1/10
9 - A = 9/10
A = 9 - 9/10
A = 81/10
ta có
\(B=\frac{1}{2}+\frac{1}{6}+................+\frac{1}{90}=\frac{1}{1x2}+.....+\frac{1}{9x10}=1-\frac{1}{2}+...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)
Dễ thấy B+A=1+1+...+1=10
=>A=10-B=\(10-\frac{9}{10}=\frac{91}{10}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+..+\left(1-\frac{1}{90}\right)\)
\(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{9.10}\right)\)
\(A=\left(1+1+1+...+1\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
9 số 1
\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)\)
\(A=9-\frac{9}{10}\)
\(A=\frac{81}{10}\)
Ủng hộ mk nha ^-^
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 .....+1 - 1/110= 10 - ( 1/2 + 1/6 + ...+ 1/110) = 10 - ( 1 - 1/ 2+ 1/2 - 1/ 3+ 1/3 - 1/4 ....+ 1/10 - 1/11)= 10 - (1 - 1/11)= 10 - 10/11 = 100/11
Sửa đề: 1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=1-1/2+1-1/6+...+1-1/90
=9-(1-1/2+1/2-1/3+...+1/9-1/10)
=9-1+1/10=8+1/10=8,1