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a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)
\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{3}-\dfrac{1}{13}\)
\(=\dfrac{10}{39}\)
nhưng khi tính trong máy tính được kết quả là \(\dfrac{5}{39}\) mà bạn
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)
=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)
=\(\dfrac{450}{119}\)
Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)
\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)
Đăt :
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.........+\dfrac{2}{49.51}\)
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..........+\dfrac{3}{49.51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.........+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=\dfrac{50}{51}\)
\(\Rightarrow A=\dfrac{50}{51}:\dfrac{3}{2}=\dfrac{100}{153}\)
Ta có công thức nha sau :
\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Ta gọi biểu thức phân số là A
Vậy \(\dfrac{2}{1.4}=\dfrac{2}{4-1}.\left(1-\dfrac{1}{4}\right)\)
\(\dfrac{2}{4.7}=\dfrac{2}{7-4}.\left(\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(\dfrac{2}{7.10}=\dfrac{2}{10-7}.\left(\dfrac{1}{7}-\dfrac{1}{10}\right)\)
Ta thấy 50 - 49 = 1 , không bằng những biểu thức kia bằng 3 nên ta tách những biểu thức đó ra.
A= \(\dfrac{2}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\right)+\dfrac{2}{49.50}\)
\(A=\dfrac{2}{3}.\left(1-\dfrac{1}{10}\right)+2.\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(A=\dfrac{18}{30}+\left(\dfrac{1}{1225}\right)=\dfrac{736}{1225}\)
mink chắc chắn, ủng hộ nha
b: \(A=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)
=1-1/16=15/16
Bài 1:
a)\(\dfrac{2\cdot\left(-3\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\dfrac{2\cdot3\cdot3\cdot2\cdot5}{2\cdot2\cdot13\cdot2}=\dfrac{45}{26}\)
b)\(\dfrac{15\cdot8+15\cdot4}{12\cdot3}=\dfrac{15\cdot12}{12\cdot13}=\dfrac{15}{13}\)
Giải:
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
Vậy ...
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{3}-\dfrac{1}{10}\)
=\(\dfrac{7}{30}\)