A=\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{50}\)
tinh A
CMR a ko la so tu nhien
Những câu hỏi liên quan
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\)
Chung minh rang A khong co gia tri la mot so tu nhien
Mk dag can gap
Đặt \(T=3\cdot5\cdot7\cdot.....\cdot49\)
\(\Rightarrow A\cdot T=\frac{T}{2}+\frac{T}{3}+\frac{T}{4}+....+\frac{T}{50}\)
\(2^4\cdot B\cdot T=\frac{2^4T}{2}+\frac{2^4T}{3}+\frac{2^4T}{4}+....+\frac{2^4T}{50}\left(1\right)\)
Tất cả các số hạng của (1) đều là stn ngoại trừ \(\frac{2^4T}{5}\)
\(\Rightarrow VP\notinℕ\Rightarrow VT\notinℕ\)
Mà \(2^4\inℕ\Rightarrow T\inℕ\)
\(\Rightarrow A\notinℕ\left(đpcm\right)\)
cmr: \(1+\frac{1}{2^2}+\frac{1}{3^2}..........+\frac{1}{100^2}\)ko phai la so tu nhien
Ta có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}>0\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}>1\) (1)
Ta lại có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
< \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
< \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
< \(1-\frac{1}{100}< 1\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)\(< 1+1\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)\(< 2\) (2)
Từ (1) và (2) => \(1< 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\)
=> \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)không là số tự nhiên
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cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap
cho s=\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
chung minh rang 1<s<2 tu do suy ra s ko phai la so tu nhien
3/10=3/9*10
3/11=3/10*11
3/12=3/11*12
3/13=3/12*13
3/14=3/13*14
suy ra 3/10+3/3/11+....+3/14 nhỏ hơn 3/9*10+....+3/13*14
suy ra 3/9*10 + 3/10*11+....+3/13*14
=1/9-1/10+....+1/13-1/14
=1/9-1/14
tự viết kết quả nhé
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cho a la so tu nhien khac khong . hay so sanh
\(\frac{a}{a+1}+\frac{a+1}{a}\) voi 2
Áp dụng bđt Cauchy cho 2 số dương \(\frac{a}{a+1}\)và\(\frac{a+1}{a}\)có
\(\frac{a}{a+1}+\frac{a+1}{a}\ge2\sqrt{\frac{a}{a+1}.\frac{a+1}{a}}=2\)
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CMR:Với moi so tu nhien n>=1thi:
a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}<\frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}=<\frac{1}{4}\)
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
< \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
< \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
< \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
< \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
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tim so tu nhien a,b,c sao cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{4}{5}\)
tim cac so tu nhien a,b,c sao cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{4}{5}\)
Chứng minh rằng số: P=\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n+1}\left(n\varepsilon N\right)\) khong la so tu nhien