Toán lớp 3?Sao giống lớp 4,5 hơn!
Ta có:
\(A=\frac{5}{15}+...+\frac{5}{399}\)
\(\Rightarrow A=5.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(\Rightarrow A=\frac{5}{7}\)
\(A=\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+...+\frac{5}{399}\)
\(A=\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+\frac{5}{7\cdot9}+...+\frac{5}{19\cdot21}\)
\(A=\frac{5}{3\cdot5\cdot2}+\frac{5}{5\cdot7\cdot2}+\frac{5}{7\cdot9\cdot2}+...+\frac{5}{19\cdot21\cdot2}\)
\(A=\frac{5}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\)
\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(A=\frac{5}{2}\cdot\left(\frac{7}{21}-\frac{1}{21}\right)\)
\(A=\frac{5}{2}\cdot\frac{6}{21}\)
\(A=\frac{30}{42}=\frac{5}{7}\)
=\(\frac{5}{3.5}\)+\(\frac{5}{5.7}\)+.......+\(\frac{5}{19.21}\)
=\(\frac{5}{2}\)(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+...........+\(\frac{2}{19.21}\))
=5/2(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+.......+\(\frac{1}{19}\)-\(\frac{1}{21}\))
=5/2(1/3-1/21)
=5/2 .6/21
=bạn tự tính nha
ta có
A=5/15+.....+5/399
suy ra A=5.(1/3.5+1/5.7+....+1/19.21
A=5/2.(1/3-1/5+...+1/19-1/21)1/3-1/21
A=5/2.(1/3-1/21)
A=5/7
AI THẤY ĐÚNG K ỦNG HỘ NH
k mik nhaKẻ Lạnh Lùng
\(A=\frac{5}{15}+\frac{5}{35}+.....+\frac{5}{399}\)
\(A=\frac{5}{3.5}+\frac{5}{5.7}+......+\frac{5}{19.21}\)
\(A=5.\left(\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\right)\)
\(A=5.\frac{2}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{19.21}\right)\)
\(A=\frac{5}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{19.21}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+........+\frac{1}{19}-\frac{1}{21}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(A=\frac{5}{2}.\frac{2}{7}\)
\(A=\frac{5}{7}\)