\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}\)
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Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
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=\(\frac{4}{13}\)
\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)
\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)
\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(=4\cdot\frac{100}{309}=\frac{400}{309}\)
NHẦM GIẢI LẠI :
\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)
D = \(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-...-\frac{1}{79.81}\)
\(=\frac{1}{54}-\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{79.81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{79.81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{81}\right)\)
\(=\frac{1}{54}-\frac{3}{2}.\frac{80}{81}\)
\(=\frac{1}{54}-\frac{40}{27}\)
\(=\frac{1}{54}-\frac{80}{54}\)
\(=\frac{79}{54}\)
S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)
S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\)
S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)
=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)
=> S = \(\frac{18}{19}x5=\frac{90}{19}\)
=3(1/1.3+1/3.5+1/5.7+1/7.9)
=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2
=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0
=3/2.8/9=4/3
ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)
ta đặt 3 làm chung rồi tự làm đc