\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)

\(\Rightarrow S=\frac{11}{20}\)

ko bao giờ 323445465

Ta có :

\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)

\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)

\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{50}{100}-\frac{1}{100}\)

\(=\frac{49}{100}\)

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 \(\frac{49}{200}\)

THANKS NHÉ!

\(S=\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}\right)-\left(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}\right).\)

Đặt A là biểu thức trong ngoặc đơn thứ nhất bà B là biểu thức trong ngoặc đơn thứ 2

\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

\(A=\frac{8}{9}:2=\frac{4}{9}\)

\(2B=\frac{4-2}{2x4}+\frac{6-4}{4x6}+\frac{8-6}{6x8}=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}\)

\(2B=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\Rightarrow B=\frac{3}{8}:2=\frac{3}{16}\)

\(S=A-B=\frac{4}{9}-\frac{3}{16}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)

\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)

k minh minh giai

Đáp số: \(\frac{49}{200}\).

Đúng 100% luôn! 

Ai tk cho mình mình tk lại.

\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

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=\(\frac{49}{200}\)

câu này mk gặp rùi nhưng ko biết cách làm

Thế thì nói làm gì?????????????

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

\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+..+\dfrac{1}{97.99}+\dfrac{1}{98.100}-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{99.100}\right)\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{98}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\dfrac{98}{99}+\dfrac{49}{100}\right]-\dfrac{49}{99}=\dfrac{14651}{19800}-\dfrac{49}{99}=\dfrac{49}{200}\)

\(\dfrac{1}{1x3}+\dfrac{1}{2x4}+...+\dfrac{1}{98x100}+\dfrac{1}{97x99}-\dfrac{49}{99}=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{100}-\dfrac{49}{99}=1-\dfrac{1}{100}-\dfrac{49}{99}\)

=\(\dfrac{4901}{9900}\)