\(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{17}-\frac{1}{31}+\frac{1}{65}-\frac{1}{127}\)Chứng minh \(A<\frac{1}{10}\)
Những câu hỏi liên quan
Cho \(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{17}-\frac{1}{31}+\frac{1}{65}-\frac{1}{127}\)
Chứng minh \(A<\frac{1}{10}\)
CMR: \(\frac{1}{5}-\frac{1}{7}+\frac{1}{17}-\frac{1}{31}+\frac{1}{65}-\frac{1}{127}< 10\)
1/5-1/7+1/17-1/331+1/65-1/127=0,01(sấp sỉ)
Vì kết quả trên nên ta kết luận:
biểu thức trên bé hơn 10
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Cho A=\(\frac{1}{5}-\frac{1}{7}+\frac{1}{17}-\frac{1}{31}+\frac{1}{65}-\frac{1}{127}.\)
So sánh A với \(\frac{1}{9}\)
bn tự giải nhé tớ gợi ý cho bn nè
gộp các số bị trừ vào 1 cặp và tất cả số trừ vào 1 cặp
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Cho Afrac{frac{1}{1.101}+frac{1}{2.102}+frac{1}{3.103}+...+frac{1}{25.125}}{frac{1}{1.26}+frac{1}{2.27}+frac{1}{3.28}+...+frac{1}{100.125}}. . Bfrac{frac{16}{9}-frac{16}{127}+frac{16}{2017}}{frac{5}{2017}+frac{5}{9}-frac{5}{127}}-frac{frac{6000}{43}-frac{6000}{257}-frac{125}{42}}{frac{2000}{43}-frac{250}{252}-frac{2000}{257}}.Chứng minh rằng Afrac{1}{2007^2}+frac{1}{2006^2}+frac{1}{2005^2}+...+frac{1}{7^2}+frac{1}{6^2}+frac{1}{5^2}B.
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Cho \(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}.\) .
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}.\)
Chứng minh rằng \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B.\)
bài này dài lắm
\(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}\)
\(A=\frac{\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)}{\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)}\)
\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}\)
\(A=\frac{\left(\frac{1}{100}\right)}{\left(\frac{1}{25}\right)}=\frac{1}{4}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}\)
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{6000}{2016}}{\frac{2000}{43}-\frac{2000}{2016}-\frac{2000}{257}}\)
\(B=\frac{16.\left(\frac{1}{9}-\frac{1}{127}+\frac{1}{2017}\right)}{5.\left(\frac{1}{2017}+\frac{1}{9}-\frac{1}{127}\right)}-\frac{6000.\left(\frac{1}{43}-\frac{1}{257}-\frac{1}{2016}\right)}{2000.\left(\frac{1}{43}-\frac{1}{2016}-\frac{1}{257}\right)}\)
\(B=\frac{16}{5}-3=\frac{1}{5}\)
Đặt \(C=\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}\)
\(C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2005^2}+\frac{1}{2006^2}+\frac{1}{2007^2}\)
\(C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}+\frac{1}{2006.2007}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2006}+\frac{1}{2006}-\frac{1}{2007}\)
\(=\frac{1}{4}-\frac{1}{2017}\left(đpcm\right)\)
\(C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2005.2006}+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}-\frac{1}{2008}\)
\(=\frac{1}{5}-\frac{1}{2008}\left(đpcm\right)\)
Vậy \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B\)
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Tìm số nguyên x
a) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
b)\(\frac{5}{17}+\frac{-9}{4}+\frac{-26}{31}+\frac{12}{17}+\frac{-11}{31}< \frac{x}{9}\le\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
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bạn ơi bạn giải câu b được ko. mk ko biết làm câu b
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\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}\)
\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)
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Bài 1:Thực hiện các phép tínha)A1-frac{1}{1+frac{2}{1-frac{3}{1-4}}}b)B-frac{1}{10}-frac{1}{100}-frac{1}{1000}-frac{1}{10000}-frac{1}{100000}-frac{1}{1000000}Bài 2:Thực hiện các phép tính sau 1 cách hợp lýa)Afrac{frac{3}{7}-frac{3}{17}+frac{3}{37}}{frac{5}{7}-frac{5}{17}+frac{5}{37}}+frac{frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}}{frac{7}{5}-frac{7}{4}+frac{7}{3}-frac{7}{2}}b)Bfrac{frac{2}{39}-frac{1}{15}-frac{2}{153}}{frac{1}{34}+frac{3}{20}-frac{3}{26}}:frac{1+frac{2}{71}-frac{5}{121}}{frac{...
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Bài 1:Thực hiện các phép tính
a)A=\(1-\frac{1}{1+\frac{2}{1-\frac{3}{1-4}}}\)
b)B=\(-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}-\frac{1}{1000000}\)
Bài 2:Thực hiện các phép tính sau 1 cách hợp lý
a)A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
b)B=\(\frac{\frac{2}{39}-\frac{1}{15}-\frac{2}{153}}{\frac{1}{34}+\frac{3}{20}-\frac{3}{26}}:\frac{1+\frac{2}{71}-\frac{5}{121}}{\frac{65}{121}-\frac{26}{71}-13}\)
c)C=\(\left(\frac{112}{13.20}+\frac{112}{20.27}+\frac{112}{27.34}+...+\frac{112}{62.69}\right):\left(-\frac{5}{9.13}-\frac{7}{9.25}-\frac{13}{19.25}-\frac{31}{19.69}\right)\)
d)D=\(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}}\)
e)E=\(-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
Bài 1 :Thực hiện phép tính :
a) M (frac{-6}{13}+frac{15}{26}-frac{47}{39}-frac{1}{78}) : (99frac{17}{65}-100frac{5}{52}+frac{1}{130})
b) N frac{(frac{3}{5}-0,435+frac{1}{200}):left(-0,04right)}{30,75+frac{1}{12}+3frac{1}{6}}
c) P (frac{-5}{6}:frac{-10}{11})+frac{frac{1}{4}+frac{5}{8}-frac{7}{13}}{frac{-2}{12}-frac{10}{24}+frac{14}{39}}
Bài 2 : Thực hiện phép tính :V
a) P frac{frac{1}{5}-frac{1}{9}+frac{1}{13}}{frac{9}{5}-1+frac{9}{13}}+frac{frac{10}{7}-frac{10}{11}-frac{10}{17}}{frac{12}{...
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Bài 1 :Thực hiện phép tính :
a) M =(\(\frac{-6}{13}+\frac{15}{26}-\frac{47}{39}-\frac{1}{78}\)) : (\(99\frac{17}{65}-100\frac{5}{52}+\frac{1}{130}\))
b) N = \(\frac{(\frac{3}{5}-0,435+\frac{1}{200}):\left(-0,04\right)}{30,75+\frac{1}{12}+3\frac{1}{6}}\)
c) P = (\(\frac{-5}{6}:\frac{-10}{11}\))+\(\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
Bài 2 : Thực hiện phép tính :V
a) P =\(\frac{\frac{1}{5}-\frac{1}{9}+\frac{1}{13}}{\frac{9}{5}-1+\frac{9}{13}}+\frac{\frac{10}{7}-\frac{10}{11}-\frac{10}{17}}{\frac{12}{7}-\frac{12}{11}-\frac{12}{17}}\)
b) Q = \(\frac{\frac{1}{14}-\frac{1}{30}-\frac{1}{46}}{\frac{2}{35}-\frac{2}{75}-\frac{2}{115}}:\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi
\(M=\frac{\frac{2}{7}-\frac{2}{13}+\frac{2}{23}}{\frac{-5}{7}+\frac{5}{13}-\frac{5}{23}}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{\frac{3}{17}-\frac{3}{23}-\frac{3}{31}}\)
\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)
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