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\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)
= 1/3 . (1/2 - 1/20)
= 1/3 . 9/20
= 3/20
\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)
\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(S=\frac{9}{20}:3=\frac{3}{20}\)
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
- A ở trên giữa các phân số là dấu " + " nha mấy bạn !
Gọi biểu thức đó là A
Ta có: \(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{17.20}\)
\(A:4.3=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(A:4.3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(A:4.3=\frac{1}{2}-\frac{1}{20}\)
\(A:4.3=\frac{9}{20}\)
\(A=\frac{3}{5}\)
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
\(a\)) Mình giải theo cách khác:
Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)
Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)
\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)
\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)
\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)
=> \(x=2\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)
\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)
\(=>x=1\)
2 phần dưới không liên quan gì đến tính chất trên
a) \(A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{20-17}{17.20}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
b) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)
\(S=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}\)
\(\Rightarrow3S=\frac{9}{20}\)
\(\Rightarrow S=\frac{3}{20}\)
\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{17\cdot20}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(S=\frac{1}{2}-\frac{1}{20}\)
\(S=\frac{9}{20}\)