Chứng minh rằng: \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
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Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}< 1\)
b) \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
a) Đặt :
\(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.................+\dfrac{1}{100!}\)
Ta thấy :
\(\dfrac{1}{2!}=\dfrac{1}{1.2}\)
\(\dfrac{1}{3!}=\dfrac{1}{1.2.3}\)
\(\dfrac{1}{4!}=\dfrac{1}{1.2.3.4}< \dfrac{1}{3.4}\)
.....................................
\(\dfrac{1}{100!}=\dfrac{1}{1.2.3..........100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
b) Đặt :
\(B=\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+.............+\dfrac{9}{1000!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{9}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
...................................................
\(\dfrac{9}{1000!}< \dfrac{1000-1}{1000!}=\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+............+\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(B< \dfrac{1}{9!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
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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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23 tháng 4 2023 lúc 22:20
\(\dfrac{1}{5}\)+\(\dfrac{1}{13}\)+\(\dfrac{1}{25}\)+...+\(\dfrac{1}{10^2}\)+\(\dfrac{1}{11^2}\)< \(\dfrac{9}{20}\)
Chứng tỏ rằng biểu thức trên bé hơn 9/20
N=\(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)
Chứng minh rằng: N<\(\dfrac{1}{9}\)
N = \(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)
= \(\dfrac{1}{10.10}+\dfrac{1}{11.11}+\dfrac{1}{12.12}+...+\dfrac{1}{n.n}\)
=> N < \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+...+\dfrac{1}{\left(n-1\right).n}\)
=> N < \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=>N< \dfrac{1}{9}-\dfrac{1}{n}\)
=> N < \(\dfrac{1}{9}\)
Vậy N < \(\dfrac{1}{9}\)
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Cho \(A=\dfrac{13}{25}+\dfrac{9}{10}-\dfrac{11}{15}+\dfrac{13}{21}-\dfrac{15}{28}+\dfrac{17}{36}-...+\dfrac{197}{4851}-\dfrac{199}{4950}\)
Chứng minh \(A>\dfrac{9}{10}\)
Chứng minh : \(\dfrac{9}{22}\) < \(\dfrac{1}{4}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...\(\dfrac{1}{100}\)<\(\dfrac{9}{10}\)
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
Chứng minh: \(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dots+\dfrac{1}{128}>2\)
B = \(\dfrac{9}{10!}+\dfrac{10}{11!}+....+\dfrac{99}{100!}\)
Chứng minh B < \(\dfrac{1}{9}\)
\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
..........................
\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
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