cho A=6/5.11+5/3.8+4/11.15-3/5.8. So sánh A với 3/10, giúp em vs ạ
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3/5 . 11/7 + 3/5 . (- 4/7) + 2/5
= 3/5 . [11/7 + (- 4/7)] + 2/5
= 3/5 . 1 + 2/5
= 3/5 + 2/5
= 1
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
mk làm phần a thui nhé
a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/2 - 1/6
A= 3/6 - 1/6
A = 1/3
\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(b=\frac{1}{2}-\frac{1}{14}\)
\(b=\frac{3}{7}\)
\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(d=1-\frac{1}{11}\)
\(d=\frac{10}{11}\)
\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
`b)`
\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)
`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)
`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)
`c)`
\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)
`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)
`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)
`d)`
\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)
`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)
`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)
a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
\(=\dfrac{7}{5}.1\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)
\(=\dfrac{-3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
\(=\dfrac{-3}{5}.2\)
\(=\dfrac{-6}{5}\)
c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)
\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)
\(=3+\dfrac{12}{5}\)
\(=\dfrac{15}{5}+\dfrac{12}{5}\)
\(=\dfrac{27}{5}\)
d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)
\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)
\(=4-\dfrac{25}{7}\)
\(=\dfrac{28}{7}-\dfrac{25}{7}\)
\(=\dfrac{3}{7}\)
Chúc bạn học tốt
A=916/55>3/10
cách làm đâu