a. \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)
b. \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)
\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
Ta có :
\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)
\(=\frac{4}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\left(\frac{11}{55}-\frac{1}{55}\right)=2.\frac{10}{55}=2.\frac{2}{11}=\frac{4}{11}\)
Có gì không hiểu bạn hỏi lại mình nhé! Chúc bạn học tốt!
Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{53.55}\)
Đặt C = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{53.55}\)
\(\frac{1}{2}C=\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+....+\left(\frac{1}{53}-\frac{1}{55}\right)\)
\(\frac{1}{2}C=\frac{1}{5}-\frac{1}{55}\)
\(\frac{1}{2}C=\frac{2}{11}\)
\(C=\frac{2}{11}:\frac{1}{2}\)
Vậy C = \(\frac{4}{11}\)
Có gì sai thì mong bạn thông cảm
\(\left(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\right)\)
Ta có :
\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+............+\frac{2}{53.55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..............+\frac{1}{53}-\frac{1}{55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\frac{2}{11}=\frac{4}{11}\)
k nha bạn !!!
1) Thực hiện phép tính:
a/ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+....+\frac{1}{97.99}\)
b/ (\(\frac{201}{202}-\frac{206}{207}+\frac{21}{199}\)) . (\(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\))
Tính ?
A.\(3+\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\)
B.\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...\frac{4}{2015.2017}\)
A. Đặt A= biểu thức đã cho
=>\(\frac{A}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=>\(\frac{A}{3}.2=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(\frac{2A}{3}-\frac{A}{3}=2-\frac{1}{2^9}\)
=>\(A=\frac{3\left(2^{10}-1\right)}{2^9}\)
B. Đặt B=biểu thức đã cho
\(\Rightarrow\frac{B}{2}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(=\frac{1}{3}-\frac{1}{2017}=\frac{2014}{6051}\)
\(\Rightarrow B=\frac{4028}{6051}\)
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
\(E=\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+...+\frac{100}{99.101}\)
a) (\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)) . x =\(\frac{1}{3}\)
b) (\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)) : x = \(\frac{2}{3}\)
c) (\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)) . x = \(\frac{2}{3}\)
Mik đang cần gấp
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)
Giống câu a