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Ta có nếu theo quy luật như trên thì sẽ có 1 thừa số là\(\frac{1}{49}-\frac{1}{7^2}\)
Mà chúng bằng 0 nên tích trên bằng 0
Gọi \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)
Từ đề bài ta có
\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)
\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)
\(D=182\times\frac{1}{8}\times\frac{80}{91}\)
\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)
cho tui nha
Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)
\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)
Vậy D=20
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
Ta có : \(\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right).......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......\left(\frac{1}{49}-\frac{1}{7^2}\right)......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......0......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=0\)