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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
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
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)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)+(1-1/90)
=(1+1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
Ta có : A=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
A=1/1.2+1/2.3+1/3.4=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=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
A=1-1/10
Thay vào ta có
=9-9/10
=81/10
Đầy đủ luôn nhé
Đặt tổng trên là A ta có :
A= 1/ 20 + 1/ 30 + 1/ 42 + 1/ 56 + 1/ 72 + 1/90 + 1/110 + 1 / 123 + 1/ 156
= 1 / 4 x5 + 1/ 5 x 6 + 1/6x 7 + 1/ 7x8 + 1/8x9 + 1/9x10+ 1/ 10x11+ 1 /11x12 +1/12 x 13
= 1/4- 1/5 + 1/ 5 - 1/6 + 1/ 6 - 1/7 + 1/ 7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10+ 1/10 - 1/11 + 1/11 - 1/12+ 1/ 12 - 1/13
= 1 /4 - 1 /13
= 9 /52
\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)
Tương tự các phân số khác
S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)
\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)
= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)
= \(\dfrac{2}{12}\)
=\(\dfrac{1}{6}\)
`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`
`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`
`=1/4-1/12=3/12-1/12=2/12=1/6`
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+......+\left(1-\frac{98}{90}\right)\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}\)
\(=\frac{81}{10}\)
(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)
=(1+1+1+1+1+1+1+1+1)-(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+...............+1/9-1/10)
=9-(1-1/10)
=9-1/9
=80/9
\(A=\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
Gọi \(A=9-B\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(A=9-\frac{9}{10}\)
\(A=\frac{90-9}{10}=\frac{81}{10}\)
Ko đúng hơi tiếc :D
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
Câu hỏi của Try Build Gundam - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}\)
\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}=\frac{2}{12}=\frac{1}{6}\)
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{110}+\frac{1}{132}\)
\(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+\frac{1}{8\times9}+...+\frac{1}{11\times12}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{4}-\frac{1}{12}=\frac{3-1}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)