Chứng minh rằng \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.............+\frac{1}{2009\sqrt{2008}}< 2\)
chứng minh rằng: \(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2008\sqrt{2007}}\)không phải là số nguyên tố
Ap dung \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Ta co \(P< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2008}}\right)\)
=> \(P< 2\left(1-\frac{1}{\sqrt{2008}}\right)< 2.1=2\)
Suy ra P khong phai so nguyen to
chứng minh rằng: \(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2008\sqrt{2007}}\)không phải là số nguyên tố
Lời giải:
Xét số hạng tổng quát $\frac{1}{(n+1)\sqrt{n}}$
Ta có:
$\frac{1}{(n+1)\sqrt{n}}=\frac{2}{2(n+1)\sqrt{n}}=\frac{2}{(n+1)\sqrt{n}+(n+1)\sqrt{n}}$
$< \frac{2}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{2(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}=\frac{2(\sqrt{n+1}-\sqrt{n})}{\sqrt{n(n+1)}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}$
Do đó:
$P< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+....+\frac{2}{\sqrt{2007}}-\frac{2}{\sqrt{2008}}=2-\frac{2}{\sqrt{2008}}< 2$
Do đó $P$ không thể là số nguyên tố.
TÍNH \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2008^2}+\frac{1}{2009^2}}\)
GIÚP E với
Tổng quát \(n\in N\text{*};n\ge2\) ta có \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)
\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\cdot1\cdot\frac{1}{n}-2\cdot1\cdot\frac{1}{n+1}-2\cdot\frac{1}{n}\cdot\frac{1}{n+1}}\)
\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\).Áp dụng vào ta có:
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2008^2}+\frac{1}{2009^2}}=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2008}-\frac{1}{2009}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)
Super dễ nhé !! Cho bn xử nốt
Chứng minh: \(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{2009}}>237\)
Bài 1. So sánh
a) \(\sqrt{2009}-\sqrt{2008}\)và \(\sqrt{2007}-\sqrt{2006}\)
b) \(\sqrt{11+\sqrt{96}}\)và \(\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
Bài 2. Tính tổng
\(T=\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)
\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)
ai cứu mk ikk
Bài 2:
\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)
Với mọi \(n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}]^2-\left(n\sqrt{n+1}\right)^2}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\left(\sqrt{n}+1\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}}{n\left(n+1\right)}-\frac{n\sqrt{n+1}}{n\left(n+1\right)}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}\)
\(=1-\frac{1}{\sqrt{121}}=\frac{10}{11}\)
Bài 1: chắc lại phải "liên hợp" gì đó rồi:V
\(\sqrt{2009}-\sqrt{2008}=\frac{1}{\sqrt{2009}+\sqrt{2008}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Đó \(\sqrt{2009}+\sqrt{2008}>\sqrt{2007}+\sqrt{2006}\)
Nên \(\sqrt{2009}-\sqrt{2008}< \sqrt{2007}-\sqrt{2006}\)
Tổng quát ta có bài toán sau, với So sánh \(\sqrt{n}-\sqrt{n-1}\text{ và }\sqrt{n-2}-\sqrt{n-3}\)
Với \(n\ge3\). Lời giải xin mời các bạn:)
Câu a)
Có: \(A=\sqrt{2009}-\sqrt{2008}\Leftrightarrow A^2=1-2\sqrt{2009\cdot2008}\)
\(B=\sqrt{2007}-\sqrt{2006}\Rightarrow B^2=1-2\sqrt{2007\cdot2006}\)
Đương nhiên: \(2\sqrt{2009\cdot2008}>2\sqrt{2006\cdot2007}\)
Suy ra: \(A< B\)
Chứng minh:
\(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{2009}}>237\)
tính b=\(1^2-2^2+3^2-...+2008^2-2009^2\)
a=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}\)
Câu a:
Có dạng tổng quát:\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{x+1}}=\frac{1}{\sqrt{\left(k+1\right)k}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{\left(k+1\right)k}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\)
Áp dụng kết quả trên suy ra câu a
Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
không dùng máy tính chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}< \frac{88}{45}\)