Tính :
A = \(\frac{17}{1.3.5}\) + \(\frac{17}{3.5.7}\) + \(\frac{17}{5.7.9}\) + ... + \(\frac{17}{47.49.51}\)
Ủng hộ 2 tick cho !!!
tính giá trị của các biểu thức
a) A= \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
b) B= \(\frac{17}{1.3.5}+\frac{17}{3.5.7}+...+\frac{17}{47.49.51}\)
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) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)
Tính tổng : \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{1997.1999.2001}\)
Tính tổng \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{1997.1999.2001}\)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)
\(=\frac{1}{1.3}-\frac{1}{1999.2001}\)
Bạn tính kết quả nhé
\(B=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
\(C=\frac{4}{9.11.13}+\frac{4}{11.13.15}+...+\frac{4}{59.61.63}\)
Ai nhanh mik tick nha !!!!!
cái này ko khó bạn áp dụng wuy luật là tính dc
B==1/4.(4/1.3.5+1/3.5.7+...+1/47.49.51)
B=1/1.3-1/3.5+1/3.5-1/5.7+....+1/47.49-1/49.50
B=1/4.(1/3.5-1/49.50)
câu B nhân lên 4 rồi tính
câu C để đó là tính dc
A = 1.3.5 + 3.5.7 + 5.7.9 + ..... + 45.47.49 + 47.49.51
chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)
Ta có:
\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)
\(\Rightarrow A=3-\frac{1}{87}\)
Vì \(3-\frac{1}{87}< 3.\)
\(\Rightarrow A< 3\left(đpcm\right).\)
Chúc bạn học tốt!
Cho A = \(\frac{1}{1.3.5}\) + \(\frac{1}{3.5.7}\) + ..... + \(\frac{1}{47.49.51}\). Chứng minh A < \(\frac{1}{12}\)
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
\(A:\)\(\frac{4}{9.11.13}+\frac{4}{11.13.15}+...+\frac{4}{59.61.63}\)
\(B:\) \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
Đây là bài tính t nha !!!
Ai nhanh mik T>I>C>K
Bài 1 : Tính nhanh : \(\frac{\frac{17}{4}-\frac{17}{16}-\frac{17}{64}+\frac{17}{256}}{13-\frac{13}{4}-\frac{13}{16}-\frac{13}{64}}\)( Trình bày rõ => tick )
khó quá tui ko biết lớp 7 à