Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}< 1\)
b) \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
Những câu hỏi liên quan
So sánh :
a) Chứng minh rằng : M = \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.......+\dfrac{1}{100!} \)
Chứng minh rằng : M <1 .
b) Chứng minh rằng : N = \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+........+\dfrac{9}{1000!}\)
Chứng minh rằng : N < \(\dfrac{1}{9!}\)
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
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Chứng minh rằng: \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
Ta có:
\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)
\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)
= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)
\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)
= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)
\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)
Chúc bn hc tốt.
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Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
Chứng minh : \(\dfrac{9}{22}\) < \(\dfrac{1}{4}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...\(\dfrac{1}{100}\)<\(\dfrac{9}{10}\)
\(a,\left(\dfrac{37}{9}+\dfrac{13}{4}\right)x\dfrac{9}{4}+\dfrac{11}{4}\) b,\(1+\left(\dfrac{9}{10}-\dfrac{-4}{5}\right):\dfrac{19}{6}\)
c,\(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}x\dfrac{12}{5}\)
Giúp mik nha:>
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
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1, Tìm các số hữu tỉ:
a) Có dạng dfrac{12}{b} sao cho dfrac{-8}{19} dfrac{12}{b} dfrac{-2}{5}
b) Có dạng dfrac{9}{b} sao cho dfrac{8}{11} dfrac{9}{b} dfrac{12}{13}
2, Tính:
M54-dfrac{1}{2}left(1+2right)-dfrac{1}{3}left(1+2+3right)-dfrac{1}{4}left(1+2+3+4right)-...dfrac{1}{12}left(1+2+3+...+12right)
3, Rút gọn các biểu thức sau:
a) A dfrac{9^9+27^7}{9^6+243^3}
b) B dfrac{left(dfrac{2}{3}right)^5.left(dfrac{-27}{8}right)^2.729}{left(dfrac{3}{2}right)^4.216}
4, Cho a,b,c là các số nguyên dư...
Đọc tiếp
1, Tìm các số hữu tỉ:
a) Có dạng \(\dfrac{12}{b}\) sao cho \(\dfrac{-8}{19}< \dfrac{12}{b}< \dfrac{-2}{5}\)
b) Có dạng \(\dfrac{9}{b}\) sao cho \(\dfrac{8}{11}< \dfrac{9}{b}< \dfrac{12}{13}\)
2, Tính:
M=\(54-\dfrac{1}{2}\left(1+2\right)-\dfrac{1}{3}\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...\dfrac{1}{12}\left(1+2+3+...+12\right)\)
3, Rút gọn các biểu thức sau:
a) A= \(\dfrac{9^9+27^7}{9^6+243^3}\)
b) B= \(\dfrac{\left(\dfrac{2}{3}\right)^5.\left(\dfrac{-27}{8}\right)^2.729}{\left(\dfrac{3}{2}\right)^4.216}\)
4, Cho a,b,c là các số nguyên dương sao cho mỗi số nhỏ hơn tổng của hai số kia. Chứng minh rằng \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\)
5, Cho A= \(\dfrac{1001}{1000^2+1}+\dfrac{1001}{1000^2+2}+...+\dfrac{1001}{1000^2+1000}\)
Chứng minh rằng 1<A2 < 4
Tính hợp lí:
A= \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}}\)
Cái này mk từng làm nhưng có chút sai sót vậy nên bn sữa cho mk chút nhé ! Thay vì N = ... thì bn thay bằng A = ... nha
Ta có :
N = 40 ( A = 40 )
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Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
19 tháng 9 2017 lúc 12:00
tinh nhanh
A=\(\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
B=\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}\)
giup minh nhe minh dang can gap
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
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