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\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)
\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)
\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)
\(=x^7.\dfrac{10}{2}\)
\(=5x^7\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)
\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)
\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)
=3/2x4/3x5/4x....x1000/999
=1/2x1000=500
mình chưa chắc là đúng đâu nhé
` a/`
` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `
`b/`
`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `
`c/`
` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `
`= 4/3 : ( 2 + 7/6 : 17/6 ) `
`= 4/3 : ( 2 + 7/6 xx 6/17 )`
`= 4/3 : ( 2 + 7/17 ) `
`= 4/3 : ( 34/17 + 7/17 ) `
`= 4/3 : 41/17 `
`= 4/3 xx 17/41 `
`= 68/123`
` d/`
` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`
\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)
= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)
= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)
= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)
= \(\dfrac{4\times25\times36}{30\times24\times35}\)
= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)
Sai rồi nhé! Từ dấu bằng thứ 2 xuống dấu bằng thứ 3 bạn làm sao đc z?
= 13/5 x 1/4 x 3/2
= 13 x 1 x 3/ 5 x 4 x 2
= 39/40
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: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
b: \(1\dfrac{1}{2}:1\dfrac{1}{3}:1\dfrac{1}{4}\)
\(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{9}{10}\)