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\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)
\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
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 = \(\frac{49}{200}\)
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=\(\frac{49}{200}\)
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
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
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 \(\frac{49}{200}\)
\(\frac{1}{2x4}\)+ \(\frac{1}{4x6}\)+ ... + \(\frac{1}{98x100}\)= \(\frac{1}{2}\)x(\(\frac{4-2}{2x4}\)+\(\frac{6-4}{4x6}\)+ ... + \(\frac{100-98}{98x100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+ ... + \(\frac{1}{98}\)-\(\frac{1}{100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{100}\)) = \(\frac{49}{200}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100
= 1/2.(2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100)
= 1/2.(1/2 - 1/4 + 1/4 - 1/6 + ... + 1/96 - 1/98 + 1/98 - 1/100)
= 1/2.(1/2 - 1/100)
= 1/2.49/100
= 49/200