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a, A= 1/5.6+1/6.7+1/7.8+...+1/24.25
=1/5-1/6+1/6-1/7+1/7-1/8+...+1/24-1/25
=1/5-1/25
=4/25
hok tốt k nha
a: \(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
b: \(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\)
\(=\dfrac{2}{15}+\dfrac{1}{8}-\dfrac{1}{87}\)
=859/3480
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9
= 1 - 1/9 = 8/9
Câu B, C dấu * là nhân hay công vậy?
\(D=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)
Áp dụng công thức tổng quát \(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
Ta có:
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\\ D=\dfrac{1}{3}-\dfrac{1}{87}\\ D=\dfrac{28}{87}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+...+\dfrac{1}{57}-\dfrac{1}{87}=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)