chứng tỏ:
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{65}\notin N\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{40}}\notin N\)
\(C=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....+\frac{1}{61}\notin N\)
A=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\). Chứng minh rằng A\(\notin\)N
Bài 1: Chứng minh rằng: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2: Cho \(n\in N;n>1\). Chứng minh rằng: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}\notin N\)
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Chứng tỏ rằng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< 1\)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< 1-\frac{1}{8}< 1\)
CHỨNG TỎ RẰNG:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}>1\)
Cho \(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Chứng minh \(E\notin Z\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Vì \(\frac{2}{6}>\frac{2}{12};\frac{2}{8}>\frac{2}{12};\frac{2}{10}>\frac{2}{12};...;\frac{1}{11}>\frac{2}{12}\)
\(\Rightarrow E=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>6.\frac{2}{12}=1\) \(\left(1\right)\)
Vì \(\frac{2}{8}< \frac{2}{6};\frac{2}{10}< \frac{2}{6};...;\frac{2}{11}< \frac{2}{6}\)
\(\Rightarrow E=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< 6.\frac{2}{6}=2\) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow1< E< 2\Rightarrow E\notin Z\)(đpcm)
Cho B = \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
CMR : B ∉ N
\(B=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{100}}\)
\(\Rightarrow B< \frac{2}{2\sqrt{1}}+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)
\(\Rightarrow B< 1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)
\(\Rightarrow B< 1+2\left(\sqrt{100}-\sqrt{1}\right)\Rightarrow B< 19\)
Tương tự:
\(B>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{101}-\sqrt{100}}\)
\(\Rightarrow B>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)
\(\Rightarrow B>2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-\sqrt{1}\right)=18\)
\(\Rightarrow18< B< 19\Rightarrow B\) không phải là số tự nhiên
Bài 1:
Chứng tỏ rằng phân số \(\frac{n+1}{2n+1}\)với n \(\varepsilon\)N và n \(\notin\)0
Bài 2:
Tìm n\(\in\)N để \(\frac{n+7}{n-2}\)\(\in\)Z
Bài 3:
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}< 1\)
b) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}< 1\)
c) \(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{37.39}>\frac{7}{13}\)
Bài 4:
Tính:
A = \(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)
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Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
Cho A = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\). CMR: A ∉ N
\(A=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+...+\frac{19}{9^2\cdot10^2}\\ A=\frac{3}{1\cdot4}+\frac{5}{4\cdot9}+...+\frac{19}{81\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)
Ta thấy \(0< \frac{99}{100}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notin N\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(\Rightarrow A=\frac{2^2-1}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(\Rightarrow A=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{10^2}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}.\)
Vì \(0< \frac{99}{100}< 1.\)
\(\Rightarrow0< A< 1.\)
\(\Rightarrow A\notin N\left(đpcm\right).\)
Chúc bạn học tốt!