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\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow\)\(x=\frac{203}{300}\)
Vậy \(x=\frac{203}{300}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}=\frac{203}{300}\)
Ta có :
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
\(=\)\(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\)
\(=\)\(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\)
\(=\)\(\frac{12}{12}-\frac{1}{100}\)
\(=\)\(1-\frac{1}{100}\)
\(=\)\(\frac{99}{100}\)
Vậy \(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
Chúc bạn học tốt ~
chắc bạn đánh thiếu đề
\(\frac{11}{1.12}+\frac{11}{12.13}+\frac{11}{23.34}+...+\frac{11}{89.100}\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-...-\frac{1}{89}+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{89.100}\)
\(=\left(1-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{24}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)\)
\(=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
đúng thì thôi. sai thì khỏi
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{13.24}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
Bài này dài dòng lắm sai thì không sai bây giờ tớ cho kết quả
= 11-11/100 = 1089/100+x =2/3
Suy ra x= 2/3 -1089/100=-3067/300(âm ba trăm sáu mươi bảy phần ba trăm)
Nếu đúng thì k ủng hộ mik nha bye chúc bạn chăm học
Sao nhiều quá vại??
mk lm k nổi đâu
Dài quá nhìn lòi bảng họng lun ak
Bài : 4
a/ \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b/ \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
c/ \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)
\(=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}+\frac{25}{26\cdot31}\)
\(=\frac{6-1}{1\cdot6}+\frac{11-6}{6\cdot11}+....+\frac{31-26}{26\cdot31}\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(=\frac{25}{5}\cdot\frac{30}{31}\)
\(=\frac{150}{31}\)
d/ \(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+....+\frac{3}{49\cdot51}\)
\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{51-49}{49\cdot51}\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\frac{50}{51}\)
\(=\frac{25}{17}\)
e/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)
\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}\)
\(=\frac{6}{37}\)
\(\left(\frac{11}{1.12}+\frac{11}{12.23}+\frac{11}{23.34}+....+\frac{11}{89.100}\right).x=\frac{1}{100}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+.....+\frac{1}{98}-\frac{1}{100}\right)x=\frac{1}{100}\)
\(\left(1-\frac{1}{100}\right).x=\frac{1}{100}\)
\(\frac{99}{100}x=\frac{1}{100}\)
\(x=\frac{1}{100}:\frac{99}{100}\)
\(x=\frac{1}{99}\)