Tìm x : 3/20-2x=1/2+5/6+11/12+19/20+29/30+...+109/110
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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
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
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
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)
\(=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{10}\right)\)
\(=\frac{91}{10}\)
11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110 =
256 phần 33
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-\(\frac{10}{11}\)=\(\frac{100}{11}\)
\(\Leftrightarrow-2x+\dfrac{3}{20}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{110}\)
\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)
\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\dfrac{10}{11}=\dfrac{100}{11}\)
=>-2x=1967/220
hay x=-1967/440