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Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ
`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`
`=1/1-1/101`
`=101/101-1/101`
`=100/101`
(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)
Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\times\left(1-\dfrac{1}{101}\right)\)
\(=2\times\dfrac{100}{101}\)
\(=\dfrac{200}{101}\)
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
Ta có : \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{49.51}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{50}\right)\)
\(=\frac{3}{2}.\frac{49}{50}=\frac{147}{100}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(Z=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{49\times51}\)
\(=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{49\times51}\right)\)
\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\times\dfrac{16}{51}=\dfrac{8}{17}\)
\(Z=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+...+\dfrac{3}{49x51}\\ =\dfrac{3}{2}x\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{49x51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\dfrac{16}{51}=\dfrac{8}{17}\)