Bài 3 : 

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)

           \(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)

            \(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)

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

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

\(A=\frac{99}{100}\)

\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)

Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)

            \(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)

           \(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)

           \(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\) 

3. 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(A=\frac{99}{100}\)

\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)

\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\)

Yến Nhi ơi !Bài B=............Ở phần mâu số là chư X hay là dấu *

9/100

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

mới đầu mình cũng định làm như lê thành đạt , nhưng x là dấu nhân thì đâu thể làm theo cách đấy .

vd:1/(10x11)x1/(11x12) khác vs 1/(10x11)+1/(11x12)

hay cậu cho đề sai ????

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

= 7/10-7/11+7/11-7/12+7/12-7/13+...+7/69-7/70

=7/10-7/70

=42/70

k mk nha

\(\frac{7}{10x11}+\frac{7}{11x12}+\frac{7}{12x13}+...+\frac{7}{69x70}\)

\(=7x\left(\frac{1}{10x11}+\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{69x70}\right)\)

\(=7x\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7x\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7x\left(\frac{7}{70}-\frac{1}{70}\right)\)

\(=7x\frac{6}{70}\)

\(=\frac{3}{5}\)

\(D=\frac{3}{3x4}+\frac{3}{4x5}+.....+\frac{3}{99x100}.\)

\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}+\frac{1}{99x100}\right)\)

\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)

\(D=1-\frac{3}{100}\)

\(D=\frac{97}{100}\)

\(D=\frac{3}{3x4}+\frac{3}{4x5}+.........+\frac{3}{98x99}+\frac{3}{99x100}\)

\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{98x99}+\frac{1}{99x100}\right)\)

\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\right)\)

\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)

\(D=\frac{3x97}{100}\)

\(D=\frac{291}{100}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+........+\frac{3}{99.100}\)

\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

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

\(=3\left(1-\frac{1}{100}\right)\)

\(=\frac{3.99}{100}=\frac{297}{100}\)