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\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\)
\(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\)
\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=19-\frac{1}{10}\)
\(=\frac{189}{10}\)
\(A=\frac{4+1}{1.2}+\frac{24+1}{3.4}+\frac{40+1}{4.5}+...+\frac{180+1}{9.10}\)
\(A=\left(\frac{4}{1.2}+\frac{24}{3.4}+\frac{40}{4.5}+...+\frac{180}{9.10}\right)+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(A=\left(2+2+2+...+2\right)+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2.8+\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{10}\right)=16+\frac{22}{30}=16\frac{11}{15}\)
\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+...+\frac{181}{9.10}\)
=\(\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+...+\frac{180+1}{90}\)
=\(2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+...+2+\frac{1}{9.10}\)
=\(18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
=\(9-\frac{1}{10}\)
=\(\frac{189}{10}\)
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 ~
Ta có
\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)
\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)
\(=-\frac{9}{10}+\frac{1}{90}\)
= ...
bn tự tính nha!
\(S=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(S=\dfrac{\left\{\left(1-2\right)^2+2.1.2\right\}}{1.2}+\dfrac{\left\{\left(2-3\right)^2+2.2.3\right\}}{2.3}+...+\dfrac{\left\{\left(9-10\right)^2+2.9.10\right\}}{9.10}\)
\(S=\dfrac{\left\{\left(-1\right)^2\right\}}{1.2+2}+\dfrac{\left\{\left(-1\right)^2\right\}}{2.3+2}+...+\dfrac{\left\{\left(-1\right)^2\right\}}{9.10+2}\)
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(S=1-\dfrac{1}{10}+18\)
\(S=\dfrac{189}{10}\)
Có sai thì đừng ném đá nha tội mình ~~
Ta có :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^2}\)
\(=\)\(\frac{3}{1.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=\)\(1-\frac{1}{100}\)
\(=\)\(\frac{100}{100}-\frac{1}{100}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Vậy ...
Đặt A=\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)
A=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.........+\frac{19}{9^2.10^2}\)
A=\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...........+\frac{19}{81.100}\)
A=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-...............+\frac{1}{81}-\frac{1}{100}\)
A=\(\frac{1}{1}-\frac{1}{100}\)
A=\(\frac{99}{100}\)
Vậy tổng của \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)là \(\frac{99}{100}\)
Chúc bn học tốt