11.Chứng tỏ :
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.................+\frac{1}{9}<2\)
Những câu hỏi liên quan
Chứng tỏ:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{10}{11}\)
Ta có:
A = \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\)+....+ \(\frac{1}{11^2}\)
A = \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + \(\frac{1}{4.4}\)+....+ \(\frac{1}{11.11}\)
A < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) + .... + \(\frac{1}{10.11}\)
A < 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ...... + \(\frac{1}{10}\) - \(\frac{1}{11}\)
A < 1 - \(\frac{1}{11}\)
\(\Rightarrow\) A < \(\frac{10}{11}\)
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Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.........
\(\frac{1}{11^2}< \frac{1}{10.11}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
Lại có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{10}{11}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{11^2}< \frac{10}{11}\) ( đpcm )
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a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
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Chứng tỏ rằng
a)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
b)\(4+2^2+2^3+2^4+.....+2^{10}=2^{11}.\)
chứng tỏ rằng :
a) \(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{10}}>\frac{1}{2^{11}}\)
b) \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}>\frac{1}{100}\)
a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}
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Chứng tỏ rằng:
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+....+\frac{1}{10^2+11^2}<\frac{9}{20}\)
Bài 1 :Chứng tỏ rằng :
a) \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b) \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
Câu hỏi của Hoàng Đỗ Việt - Toán lớp 6 | Học trực tuyến
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Bài 1 :
Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)
Mặt khác :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)
Từ (1 ) và (2) ta suy ra điều phải chứng minh
Bài 2 :
Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
MỘT MẶT ,TA CÓ THỂ VIẾT
\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)
\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)
Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)
Từ (1) và (2 ) ta kết luận \(3< S< 6\)
Chúc bạn học tốt ( -_- )
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a) Đặt \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}\)
Ta có:
\(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)
+ Vì \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};...;\frac{1}{40}=\frac{1}{40}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}.\)
+ Vì \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{60}=\frac{1}{60}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( 20 phân số \(\frac{1}{60}\)) \(=20.\frac{1}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}=\frac{75}{90}>\frac{66}{90}=\frac{11}{15}\)
\(\Rightarrow A>\frac{11}{15}\left(1\right)\)
Lại có:
+ Vì \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...;\frac{1}{40}< \frac{1}{20}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( 20 phân số \(\frac{1}{20}\)) \(=20.\frac{1}{20}=1\)
+ Vì \(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{60}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}\)
\(\Rightarrow A< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< \frac{3}{2}\left(2\right)\)
Từ\(\left(1\right)\);\(\left(2\right)\) \(\Rightarrow\frac{11}{15}< A< \frac{3}{2}\left(đpcm\right).\)
b) Đặt \(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
Ta có:
\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)\)\(+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)
+ \(1=1\)
+\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)
+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)
Tương tự ta được:
+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< 1\)
+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< 1\)
\(\Rightarrow A< 1+1+1+1+1+1=6\left(1\right)\)
Lại có:
+\(1=1\)
+\(\frac{1}{2}+\frac{1}{3}>\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{4}{7}\)
+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}>\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}=\frac{8}{15}\)
Tương tự, ta được:
+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}>\frac{16}{31}\)
+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< \frac{32}{63}\)
\(\Rightarrow A>1+\frac{2}{3}+\frac{4}{7}+\frac{8}{15}+\frac{16}{31}+\frac{32}{63}\)\(=1+\frac{18}{15}+\frac{64}{63}+\frac{16}{31}>1+\frac{15}{15}+\frac{63}{63}=3\left(2\right)\)
Từ\(\left(1\right)\)và \(\left(2\right)\Rightarrow3< A< 6\left(đpcm\right).\)
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frac{1}{3}-frac{3}{5}+frac{5}{7}-frac{7}{9}+frac{9}{11}-frac{11}{13}+frac{11}{15}+frac{11}{13}-frac{9}{11}+frac{7}{9}-frac{5}{7}+frac{3}{5}-frac{1}{3}frac{-3}{4}.31frac{11}{23}-0,75.8frac{12}{28}left(2frac{1}{3}+3frac{1}{2}right):left(-4frac{1}{6}+3frac{1}{7}right)+7frac{1}{2}4frac{5}{9}:left(-frac{5}{7}right)+5frac{4}{9}:left(-frac{5}{7}right)frac{2}{3}-4left(frac{1}{2}+frac{3}{4}right)left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)....left(1-frac{1}{n+1}right)nin Z
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\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)\(\frac{-3}{4}.31\frac{11}{23}-0,75.8\frac{12}{28}\)\(\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)\(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)\(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)\)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{n+1}\right)n\in Z\)
1.Cho A frac{5n-11}{n-2}left(ninℤright)a. Tìm điều kiện n để A là phân số b.Tìm n inℤđể A có giá trị nguyên c.Tìm giá trị lớn nhất của A2.Tìm xa. frac{3}{4}timesleft(frac{1}{2}x+frac{1}{3}right)-frac{1}{2}frac{2}{3}x-frac{1}{4}b.frac{2}{3}x-3x+frac{1}{5}frac{3}{2}left(x-frac{1}{4}right)-frac{3}{2}3.a.Chứng tỏ :frac{1}{7^2}+frac{1}{8^2}+..................+frac{1}{99^2} frac{1}{6}b.Chứng tỏ:frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+frac{1}{7}+frac{1}{8}+frac{1}{9}+frac{1}{10}+frac{1}{11}+frac{1}...
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1.Cho A =\(\frac{5n-11}{n-2}\left(n\inℤ\right)\)
a. Tìm điều kiện n để A là phân số
b.Tìm n \(\inℤ\)để A có giá trị nguyên
c.Tìm giá trị lớn nhất của A
2.Tìm x
a. \(\frac{3}{4}\times\left(\frac{1}{2}x+\frac{1}{3}\right)-\frac{1}{2}=\frac{2}{3}x-\frac{1}{4}\)
b.\(\frac{2}{3}x-3x+\frac{1}{5}=\frac{3}{2}\left(x-\frac{1}{4}\right)-\frac{3}{2}\)
3.a.Chứng tỏ :
\(\frac{1}{7^2}+\frac{1}{8^2}+..................+\frac{1}{99^2}< \frac{1}{6}\)
b.Chứng tỏ:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+............+\frac{1}{23}< 3\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Bạn tham khảo nhé
\(a)\)Đặt \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm )
Vậy \(A< 1\)
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