Đây là bài tính t nha !!!

Ai nhanh mik T>I>C>K

Bài làm:

Ta có: \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)

\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{12}-\frac{1}{4.49.51}< \frac{1}{12}\)

Vậy \(A< \frac{1}{12}\)

Từ đề bài suy ra\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{49.51}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{12}\left(đpcm\right)\)

\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)

\(4A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\)

\(4A=\frac{1}{1.3}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{2499}< \frac{1}{3}\)

=>    \(A< \frac{1}{3}:4=\frac{1}{12}\)

VẬY TA CÓ ĐPCM.

NẾU ĐÚNG MONG MỌI NGƯỜI ỦNG HỘ Ạ

=1/1.3-1/3.5+1/3.5-1/5.7+...+1/99.11-1/11.13

=1/1.3-1/11.13

=1/3-1/143

=140/429

\(\begin{equation} x = a_0 + \cfrac{1}{740_1 + \cfrac{1}{897654_2 + \cfrac{1}{672_3 + \cfrac{1}{100_4} } } } \end{equation}\)

a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)

\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)

\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)

b = 1

a đg nghĩ

@32526313:

Giải bài đầy đủ ra đi bạn =))

Nói đ/a ai chả nói được (:

Trả lời:

\(a,\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)

\(b,\frac{1991}{1990}\times\frac{1992}{1991}\times\frac{1993}{1992}\times\frac{1994}{1993}\times\frac{995}{997}\)

\(=\frac{1991\times1992\times1993\times1994\times995}{1990\times1991\times1992\times1993\times997}\)

\(=\frac{1994.995}{1990.997}=\frac{997.2.995}{995.2.997}=1\)

\(=\frac{1}{4}.\left(\frac{17.4}{1.3.5}+\frac{17.4}{3.5.7}+\frac{17.4}{5.7.9}+...+\frac{17.4}{47.49.51}\right)\)

\(=\frac{17}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)

\(=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}.\frac{832}{2499}=\frac{208}{147}\)

\(A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{25.27.29}\right)\)

\(A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)

\(A=9.\frac{260}{87}=\frac{260}{87}