\(\frac{1}{2x4}\)+ \(\frac{1 }{4x6}\)+ \(\frac{1}{6x8}\) + \(\frac{1}{8x10}\)+ ........... + \(\frac{1}{98x100}\) = ?
A=\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+......+\frac{1}{98x100}\)
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Vậy: \(A=\frac{49}{100}\)
Ta có:\(2A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{98.100}\right)\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của A là \(\frac{49}{200}\)
\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+\frac{1}{8x10}+\frac{1}{10x12}\)
các bạn làm cả lời giải giúp mk nhé!
=1/2-1/4+1/4-1/6+1/61/8+1/81/10+1/10-1/12
=1/2-1/12
5/12
tk nha
Đặt : \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{10}-\frac{1}{12}\)
\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow A=\frac{5}{12}\)
bài duoi giải sai nha
Dặt biểu thức trên là A
ta có\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{12}\)
\(2A=1-\frac{1}{12}=\frac{11}{12}\)
\(A=\frac{11}{12}\)/ 2=\(\frac{11}{24}\)
Tìm y biết \(\left(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+\frac{1}{8x10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).y=\frac{1}{3}\)
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}:\frac{1}{2}=\frac{2}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{2}{3}\)
\(\frac{2}{5}.y=\frac{2}{3}\)
=> \(y=\frac{2}{3}:\frac{2}{5}\)
=>\(y=\frac{3}{5}\)
\(S=\frac{1}{1x3}-\frac{1}{2x4}+\frac{1}{3x5}-\frac{1}{4x6}+\frac{1}{5x7}-\frac{1}{6x8}+\frac{1}{7x9}-\frac{1}{8x10}\)
giúp mình với mình đang cần gấp.
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+.............+\frac{1}{96x98}+\frac{1}{98x100}\)
Ai trả lời thì ghi cả cách làm giúp mình với nha ! Thank you
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)
Đáp số: \(\frac{49}{200}\).
Đúng 100% luôn!
Ai tk cho mình mình tk lại.
Tính:
\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+..................+\frac{1}{96x98}+\frac{1}{98x100}\)
Nếu giải đúng có lời giải ,sẽ có 2 tick
\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
A x2 = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{100}\)
A x2 = \(\frac{49}{100}\)
A = \(\frac{49}{200}\)
\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+......+\frac{1}{96x98}+\frac{1}{98x100}.\)
Chú giải: x là dấu nhân
Giải thích mình sẽ **** !!!!!
Đặt A=\(\frac{1}{2x4}+\frac{1}{4x6}+.........+\frac{1}{98x100}\)
2A=\(\frac{2}{2x4}+\frac{2}{4x6}+.............+\frac{2}{98x100}\)
2A=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..........+\frac{1}{98}-\frac{1}{100}\)
2A=\(\frac{1}{2}-\frac{1}{100}\)
2A=\(\frac{49}{100}\)
A=\(\frac{49}{100}:2\)
A=\(\frac{49}{200}\)
câu này mk gặp rùi nhưng ko biết cách làm
Tính:
\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+...........+\frac{1}{96x98}+\frac{1}{98x100}\)
giải nhanh ,có lời giải cơ sở sẽ có tick
1/2*(2/2*4+2/4*6+...+2/98*100)=1/2*(1/2-1/4+1/4-1/6+...+1/98-1/100)
=1/2*(1/2-1/100)
=1/2*49/100
=49/200
k nha bạn
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
cho mình tròn 1550 nhé bạn
=2/2 4+2/4 6+2/6 8+...+2/96 98+2/98 100
=2/2-2/4+2/4-2/6+2/6-2/8+...+2/96-2/98+2/98-2/100
=2/2-2/100
=99/100
\(\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+\frac{4}{8x10}+...+\frac{4}{16x18}+\frac{4}{18x20.}\)
Ta gọi biểu thức đó là A
Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức ta có
\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)
\(....................\)
\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)
Ai thấy đúng thì ủng hộ nha !!!
sai rồi kết quả phải bằng 9/10 chứ
\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{18\cdot20}\)
\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=2\left(\frac{9}{20}\right)\)\(=\frac{18}{20}=\frac{9}{10}\)