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Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
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:)
\(\frac{1}{\sqrt{2009}-\sqrt{2008}}=\frac{\sqrt{2009}+\sqrt{2008}}{\left(\sqrt{2009}+\sqrt{2008}\right)\left(\sqrt{2009}-\sqrt{2008}\right)}=\frac{\sqrt{2009}+\sqrt{2008}}{2009-2008}=\sqrt{2009}+\sqrt{2008}\)
CMTT : \(\frac{1}{\sqrt{2008}-\sqrt{2007}}=\sqrt{2008}+\sqrt{2007}\)
Vì \(\sqrt{2009}+\sqrt{2008}>\sqrt{2008}+\sqrt{2007}\)
=> \(\frac{1}{\sqrt{2009}-\sqrt{2008}}\sqrt{2008}-\sqrt{2007}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{\sqrt{2007}-\sqrt{2006}}{2007-2006}=\frac{\sqrt{2007}-\sqrt{2006}}{\left(\sqrt{2007}-\sqrt{2006}\right)\left(\sqrt{2007}+\sqrt{2006}\right)}\)
\(=\frac{1}{\sqrt{2007}+\sqrt{2006}}< \frac{1}{\sqrt{2006}+\sqrt{2006}}=\frac{1}{2\sqrt{2006}}\)
Vậy \(\sqrt{2007}-\sqrt{2006}< \frac{1}{2\sqrt{2006}}\)
Bạn áp dùng biểu thức liên hợp là được
Ta có :
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)(1)
\(\frac{1}{2\sqrt{2006}}=\frac{1}{\sqrt{2006}+\sqrt{2006}}\)(2)
Từ (1)(2)=>\(\frac{1}{\sqrt{2007}+\sqrt{2006}}< \frac{1}{\sqrt{2006}+\sqrt{2006}}\)
\(\Rightarrow\sqrt{2007}-\sqrt{2006}>\frac{1}{2\sqrt{2006}}\)
\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)
Vậy PT có nghiệm \(x=2008\)
Ta có
\(\hept{\begin{cases}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\left(1\right)\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\left(2\right)\end{cases}}\)
\(\Rightarrow\frac{1}{\sqrt{2008}+\sqrt{2005}}+\frac{1}{\sqrt{2010}+\sqrt{2007}}>\frac{2}{\sqrt{2015}+\sqrt{2009}}\)
\(\Leftrightarrow\frac{\sqrt{2008}-\sqrt{2005}}{3}+\frac{\sqrt{2010}-\sqrt{2007}}{3}>\frac{\sqrt{2015}-\sqrt{2009}}{3}\)
\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)
\(A=\sqrt{2007}-\sqrt{2006}=\frac{\left(\sqrt{2007}-\sqrt{2006}\right)\left(\sqrt{2007}+\sqrt{2006}\right)}{\left(\sqrt{2007}+\sqrt{2006}\right)}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)(1)
\(B=\sqrt{2008}-\sqrt{2007}=\frac{\left(\sqrt{2008}-\sqrt{2007}\right)\left(\sqrt{2008}+\sqrt{2007}\right)}{\left(\sqrt{2008}+\sqrt{2007}\right)}=\frac{1}{\sqrt{2008}+\sqrt{2007}}\)(2)
Từ 1 và 2 => \(\frac{1}{\sqrt{2007}+\sqrt{2006}}>\frac{1}{\sqrt{2008}+\sqrt{2007}}\)
hay \(\sqrt{2007}-\sqrt{2006}>\sqrt{2008}-\sqrt{2007}\)
P/s tham khảo nha