3/2.( 2/ 10.12+2/12.14+.....+2/48.50)

3/2.(1/10-1/12+1/12-1/14+....+1/48-1/50)

3/2.(1/10-1/50)

3/2.2/25
3/25

Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{10}-\frac{1}{50}\)

\(=\frac{2}{25}\)

=> 2/25

Ta có:

\(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\)

\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{82}-\frac{1}{87}\right)\)

\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{87}\right)=\frac{3}{5}.\frac{85}{174}=\frac{17}{58}\)

Ta có: 

A = \(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\)( dấu chấm là dấu nhân nhá, lên lớp 6 thì bạn sẽ biết )

\(\frac{5}{3}.A=\frac{5}{3}.\left(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{82.87}\right)\)

\(\frac{5}{3}.A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{82.87}\)

\(\frac{5}{3}.A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}+\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{82}-\frac{1}{87}\)

\(\frac{5}{3}.A=\frac{1}{2}-\frac{1}{87}\)

\(\frac{5}{3}.A=\frac{85}{174}\)

\(A=\frac{85}{174}:\frac{5}{3}=\frac{17}{58}\)

                                             Đ/s : \(\frac{17}{58}\)

Ủng hộ mik nhá !!!

3.1/2.7+3.1/7.12+3.1/12.17+.......+3.1/82.87

= 1/2-1/7+1/7-1/12+1/12-1/17+........+1/83 - 1/87

=1/2-1/87

=85/174

dấu gạch chéo thay cho gạch ngang phân số bạn nhé

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{97}{48^2.49^2}+\frac{99}{49^2.50^2}\)

\(\Leftrightarrow\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{97}{2304.2401}+\frac{99}{2401.2500}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{2304}-\frac{1}{2401}+\frac{1}{2401}-\frac{1}{2500}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2500}=\frac{2499}{2500}< 1\left(đpcm\right)\)

Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
 1) Ta có:    \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)

                                                                             \(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)

                                                                             \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}

=1-1/50=\(\frac{49}{50}\)

Tách: 1/1.2=1-1/2;  1/2.3=1/2-1/3; ....; 1/49.50=1/49-1/50

Và rút gọn các số liền kề thì còn lại kết quả

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(=\left(1-\frac{1^1}{2}\right)+\left(\frac{1^1}{2}-\frac{1^1}{3}\right)+\left(\frac{1^1}{3}-\frac{1}{4}^1\right)+...+\left(\frac{1^1}{49}-\frac{1}{50}\right)\)

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

\(=\frac{49}{50}\)

\(\frac{45+45}{300+600}\)

\(\frac{90}{900}\)

\(\frac{1}{10}\)

300600

Tui là người trả lời đầu tiên đó nhoa

\(B=\frac{\left(2^4.3\right)^3.\left(2.5^2\right)^5}{\left(2^6\right)^2.\left(5^3\right)^2.\left(2.3.5\right)^3}\)

\(B=\frac{2^{12}.3^3.2^5.5^{10}}{2^{12}.5^6.2^3.3^3.5^3}\)

\(B=\frac{2^{12}.3^3.2^3.2^2.5^9.5}{2^{12}.5^9.2^3.3^3}\)

\(B=2^2.5=20\)

Vay B = 20