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\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)=9-1+\frac{1}{10}=8\frac{1}{10}\)
A=1-1/2+1-1/6+...+1-1/90
=9-(1/2+1/6+...+1/90) =9-(1/1.2+1/2.3+...+1/9.10)
=9-(1-1/10)=9-9/10=81/10
\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
=>\(A=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
=>\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=>\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=>\(a=9-\left(1-\frac{1}{10}\right)=\frac{90}{10}-\frac{9}{10}=\frac{81}{10}\)
9 - A = \(1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+..+1-\frac{89}{90}=\frac{1}{2}+\frac{1}{6}+..+\frac{1}{90}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
=> A = \(9-\frac{9}{10}=\frac{81}{10}\)
Q=\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{89}{90}\)
Q=\(Q=\frac{5}{2.3}+\frac{11}{3.4}+\frac{19}{4.5}+\frac{29}{5.6}+\frac{41}{6.7}+\frac{55}{7.8}+\frac{71}{8.9}+\frac{89}{9.10}\)
Q=
\(A=\)\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=\)\(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(A=\)\(9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\)\(9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\)\(9-\left(1-\frac{1}{10}\right)\)
\(A=\)\(9-\frac{9}{10}\)
\(A=\)\(\frac{81}{10}\)
A=(1-1/2)+(1-1/6)+...+(1-89/90)
A=1x9-(1/2+1/6+...+1/90)
A=9-(1/1x2+1/2x3+...+1/9x10)
A=9-(1-1/2+1/2-1/3+1/3+...+1/9 -1/10)
A=9-(1-1/10)
A=9-9/10
A=81/10=8,1
hok tốt nhé
Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
\(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
mk chỉ cần nhìn sơ qua là biết có câu dễ sao bn ko tự nghĩ đi hơi dễ rồi trừ khi bn đố tôi chục câu tiếng anh vật lí văn
=> A = ( 1 - 1/2 ) + ( 1 - 1/6 ) + ( 1 - 1/12 ) + ( 1 - 1/30 ) + .... + ( 1 - 1/90 )
=> A = ( 1 + 1 + 1 + .... 1 ) - ( 1/2 + 1/6 + 1/12 + 1/30 + .... + 1/90 )
=> A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/9.10 )
=> A = 9 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10 )
=> A = 9 - ( 1 - 1/10 )
=> A = 9 - 9/10
=> 81/10
81/10