Chứng minh rằng: \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+\(\frac{1}{\sqrt{3}}\)+......+\(\frac{1}{\sqrt{121}}\)>11
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\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{121}-\sqrt{120}}{\left(\sqrt{121}-\sqrt{120}\right)\left(\sqrt{121}+\sqrt{120}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{121}-\sqrt{120}\)
\(=\sqrt{121}-1=10\)
\(B=\frac{2}{2.\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B>\frac{1}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)
\(B>2\left(\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{\sqrt{36}-\sqrt{35}}{\left(\sqrt{36}-\sqrt{35}\right)\left(\sqrt{36}+\sqrt{35}\right)}\right)\)
\(B>2\left(\sqrt{36}-\sqrt{1}\right)=10\Rightarrow B>A\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
\(\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{121}}\)
\(\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{121}}\)
................
\(\frac{1}{\sqrt{121}}=\frac{1}{\sqrt{121}}\)
Suy ra \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+.............+\(\frac{1}{\sqrt{121}}\)<\(\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+......\frac{1}{\sqrt{121}}\)=\(\frac{121}{11}\)=11(đpcm)(vì có 121 chữ số)\(\frac{1}{\sqrt{121}}\))
Khuyển Dạ Xoa : \(\sqrt{1}< \sqrt{121}\Rightarrow\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{121}}\) chứ?