Tinh
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{90}+\frac{1}{100}\)
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1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
Bạn sai đè thì phải,đúng phải là 1/99
Ta thấy:Từ 1->1/100 có 100 số.
Ta có:100=1.100
Vì 1=1 ;1/2<1 ;1/3<1 ;1/4<1 ;... ;1/90<1 ;1/100<1.
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 1.100=100\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 100\)
M=100
Xét tử N
92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)
=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)
=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]
=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]
=>N=40
=>M/N=5/2
-149/50
49/50
tích nhoa