A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

sai ùi 

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}.\frac{9}{20}\)

\(A=\frac{3}{20}\)

Vì \(\frac{3}{20}< 1\Rightarrow A< 1\)

Ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)

B=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}+\frac{3}{17.20}\)

\(\Rightarrow B=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{20}=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)

\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

=\(\frac{3}{20}=0,15\)

\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

- A ở trên giữa các phân số là dấu " + " nha mấy bạn !

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}\)

\(=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}\)

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(A=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{21}\)

\(A=\frac{1}{3}-\frac{1}{21}\)

\(A=\frac{2}{7}\)

\(B=\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{68.71}\)

\(=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+...+\frac{1}{68}-\frac{1}{71}\)

\(=\frac{1}{11}-\frac{1}{71}\)

\(=\frac{71}{781}-\frac{11}{781}\)

\(=\frac{60}{781}\)

\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)

\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)

\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)

\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)

=> \(x=2\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)

\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)

\(=>x=1\)

\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)

\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=3.\left(1-\frac{1}{10}\right)\)

\(A=3.\frac{9}{10}=\frac{27}{10}\)

\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)

\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)

a) Lấy A chia 3

b) Lấy B nhân 3/2

15544525 nfhg

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) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)

Có liên quan đó bạn!