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\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
A=1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+...+1/(10*11)
A=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/10-1/11
A=1/2-1/11
A=9/22
Đặt tổng trân là A
Ta có A=1/2x3+1/3x4+...+1/10x11
A=1/2-1/3+1/3-1/4+...+1/10-1/11
Sau khi rút gọn ta được A=1/2-1/11
A=9/22
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
=1/2.3+1/3.4+1/4.5+......+1/10.11
=1-1/2+1/2-1/3+.....,+1/10-1/11
=1-1/11
=10/11
Tick
Đặt biểu thức là A. A có 10 số hạng.
A = 1/2+5/6+11/12+19/20+...+109/110.
A = (1-1/2) + (1-1/6) + ...+(1-1/110)
A = 1+1+1+...+1(10 số 1) - (\(\frac{1}{2}\)+\(\frac{1}{6}\)+...+\(\frac{1}{110}\))
A=10-B
B = \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{10.11}\)
B = \(\frac{2-1}{1.2}\)+\(\frac{3-2}{2.3}\)+\(\frac{4-3}{3.4}\)+...+\(\frac{11-10}{10.11}\)
B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{10}\)-\(\frac{1}{11}\)
B=1-\(\frac{1}{11}\)=\(\frac{10}{11}\)
⇒A=10-B=10-\(\frac{10}{11}\)=\(\frac{100}{11}\)
\(A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\left(\frac{1}{2}-\frac{1}{11}\right)+0+...+0\)
\(A=\frac{11}{22}-\frac{2}{22}\)
\(A=\frac{9}{22}\)
Sửa đề
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
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
`1/2+1/6+1/12+1/20+...+1/110`
`= 1/(1 . 2)+1/(2 . 3)+1/(3 . 4)+1/(4.5)+...+1/(10.11)`
`= 1-1/2+1/2-1/3+1/3-1/4+1/4-...-1/10+1/10-1/11`
`= 1-1/11`
`=11/11-1/11`
`= 10/11`
10/11