a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)

\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)

b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(=\frac{15}{16}\)

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\(a)\) \(A=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{49.51}\)

\(A=3\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.50}\right)\)

\(2A=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.50}\right)\)

\(2A-A=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=3\left(\frac{1}{3}-\frac{1}{50}\right)\)

\(A=1-\frac{3}{50}\)

\(A=\frac{47}{50}\)

Vậy \(A=\frac{47}{50}\)

\(b)\) \(B=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(B=1-\frac{1}{16}\)

\(B=\frac{15}{16}\)

Vậy \(B=\frac{15}{16}\)

Chúc bạn học tốt ~

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

\(B=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}\left(\frac{1}{7}-\frac{1}{11}\right)+....+\frac{1}{4}\left(\frac{1}{107}-\frac{1}{111}\right)\)

  \(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{107}-\frac{1}{111}\right)\)

  \(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{111}\right)\)

  \(=\frac{107}{444}\)

vậy biểu thức trên =\(\frac{107}{444}\)

xong zùi đó bn ak

B=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(B=\frac{1}{3}-\frac{1}{111}\)

\(B=\frac{111}{333}-\frac{3}{333}\)

\(B=\frac{108}{333}\)

\(B=\frac{12}{37}\)

Vậy B =\(\frac{12}{37}\)

\(\left[36\times4-4\times\left(82-7\times11\right)^2\right]:4-2022^0\\ =\left[144-4\times\left(82-77\right)^2\right]:4-1\\ =\left[144-4\times5^2\right]:4-1\\ =\left[144-4\times25\right]:4-1\\ =\left[144-100\right]:4-1\\ =44:4-1=11-1=10\)

= 10

a=4/3.7 +4/7.11+4/11.15 +.....+4/107/111

=1/3-1/7+1/7-1/11+1/11-1/15+......+1/107-1/111

=1/3-1/111

=12/37

12/37nhe

Đặt A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{103.107}+\frac{4}{107.111}\)

\(=\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right)+...+\left(\frac{1}{103}-\frac{1}{107}\right)+\left(\frac{1}{107}-\frac{1}{111}\right)\)

\(=\frac{1}{3}-\frac{1}{111}=\frac{37-1}{111}=\frac{36}{111}\)

a) 1/(5.7) + 1/(7.9) + ... + 1/(2011.2013)

= 1/2.(1/5 - 1/7 + 1/7 - 1/9 + ... + 1/2011 - 1/2013)

= 1/2.(1/5 - 1/2013)

= 1/2 . 2008/10065

= 1004/10065

b) 1/(7.11) + 1/(11.15) +1/(15.19) + ... + 1/(2019.2023)

= 1/4.(1/7 - 1/11 + 1/11 - 1/15 + 1/15 - 1/19 + ... + 1/2019 - 1/2023)

= 1/4.(1/7 - 1/2023)

= 1/4 . 288/2023

= 72/2023

\(\dfrac{2^7.9^4}{4^4.3^9}\)

\(=\dfrac{2^7.\left(3^2\right)^4}{\left(2^2\right)^4.3^9}\)

\(=\dfrac{2^7.3^8}{2^8.3^9}\)

\(=\dfrac{1}{2.3}=\dfrac{1}{6}\)

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