A = 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110=?
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\(A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{9}{10.11}+\frac{1}{10}-1\)
\(=-\frac{9}{11}\)
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=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
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
Ta viết lại biểu thức A như sau:
\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)
\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(A=-\dfrac{1}{6}\)
ta có: 1/2=1/1x2
1/6=1/2x3
1/12=1/3x4
1/20=1/4x5
1/30=1/5x6
1/42=1/6x7
1/56=1/7x8
1/72=1/8x9
1/90=1/9x10
1/110=1/10x11
tiếp theo bn tiếp tục nhé
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110
=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x1
=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+1/10-1/11
=1-1/11
=10/11
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}=\frac{4}{12}-\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
Chú ý: \(\cdot=\times\)
Đặt \(A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{12}=\frac{1}{4}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)
\(A=\frac{11}{22}-\frac{2}{22}\)
\(A=\frac{9}{22}\)