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\(S=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(S=\dfrac{\left\{\left(1-2\right)^2+2.1.2\right\}}{1.2}+\dfrac{\left\{\left(2-3\right)^2+2.2.3\right\}}{2.3}+...+\dfrac{\left\{\left(9-10\right)^2+2.9.10\right\}}{9.10}\)
\(S=\dfrac{\left\{\left(-1\right)^2\right\}}{1.2+2}+\dfrac{\left\{\left(-1\right)^2\right\}}{2.3+2}+...+\dfrac{\left\{\left(-1\right)^2\right\}}{9.10+2}\)
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(S=1-\dfrac{1}{10}+18\)
\(S=\dfrac{189}{10}\)
Có sai thì đừng ném đá nha tội mình ~~
Câu b, B=\(\dfrac{5}{1\cdot2}+\dfrac{13}{2\cdot3}+\dfrac{25}{3\cdot4}+...+\dfrac{181}{9\cdot10}\)
\(=\left(\dfrac{1}{1\cdot2}+\dfrac{4}{1\cdot2}\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{12}{2\cdot3}\right)+\left(\dfrac{1}{3\cdot4}+\dfrac{24}{3\cdot4}\right)+...+\left(\dfrac{1}{9\cdot10}+\dfrac{180}{9\cdot10}\right)\)=\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)+\left(\dfrac{4}{1\cdot2}+\dfrac{12}{2\cdot3}+...+\dfrac{180}{9\cdot10}\right)\)
=\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)\(+\left(2+2+2+.......+2\right)\)
=\(\dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-......-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\dfrac{1}{10}+\left(2\cdot9\right)\)
=\(1-\dfrac{1}{10}+18\) \(=\dfrac{9}{10}+18\)
=18.9
a, \(\dfrac{\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}}{\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}}=\dfrac{\dfrac{15}{10}-\dfrac{4}{10}+\dfrac{1}{10}}{\dfrac{18}{12}-\dfrac{8}{12}+\dfrac{1}{12}}=\dfrac{\dfrac{15-4+1}{10}}{\dfrac{18-8+1}{12}}=\dfrac{\dfrac{12}{10}}{\dfrac{11}{12}}=\dfrac{72}{55}\)
\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\)
\(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\)
\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=19-\frac{1}{10}\)
\(=\frac{189}{10}\)
Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
đặt \(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
ta có:
\(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
\(\Leftrightarrow M=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}\) \(\Leftrightarrow M=\dfrac{3}{3.4}+\dfrac{4}{3.4}-\dfrac{4}{4.5}-\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{6}{5.6}-\dfrac{6}{6.7}-\dfrac{7}{6.7}+\dfrac{7}{7.8}+\dfrac{8}{7.8}-\dfrac{8}{8.9}-\dfrac{9}{8.9}+\dfrac{9}{9.10}+\dfrac{10}{9.10}\) \(\Rightarrow M=\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{10}+\dfrac{1}{9}\) \(\Rightarrow M=\dfrac{1}{3}+\dfrac{1}{10}\)
\(\Rightarrow M=\dfrac{10}{30}+\dfrac{3}{30}\)
\(\Rightarrow M=\dfrac{13}{30}\)
vậy M = \(\dfrac{13}{30}\)
vậy \(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{13}{30}\)
\(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{7}{30}\)
C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)
= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2
= 1/9 - 1/10 + 1/9 - 1
= 2/9 - 11/10
= -79/90
Đầu tiên thì nhắc lại cái hằng đẳng thức cho bạn nào chưa học này: (a-b)2=a2-2ab+b2<=>a2+b2=(a-b)2+2ab
\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(=\dfrac{\left(\left(1-2\right)^2+2.1.2\right)}{1.2}+\dfrac{\left(\left(2-3\right)^2+2.2.3\right)}{2.3}+...+\dfrac{\left(\left(9-10\right)^2+2.9.10\right)}{9.10}\)
\(=\dfrac{\left(\left(-1\right)^2\right)}{1.2+2}+\dfrac{\left(\left(-1\right)^2\right)}{2.3+2}+...+\dfrac{\left(\left(-1^2\right)\right)}{9.10+2}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(=1-\dfrac{1}{10}+18\)
\(=18,9=\dfrac{189}{10}.\)
~ K chắc là đúng đâu ~