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Ta co:
\(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}.\sqrt{n}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Ap vào bài toan được
\(S_n=\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{n+1}}\right)< \frac{1}{2}\)
\(P=\dfrac{\sqrt{2x^2+y^2}}{xy}+\dfrac{\sqrt{2y^2+z^2}}{yz}+\dfrac{\sqrt{2z^2+x^2}}{xz}\)
\(P=\sqrt{\dfrac{2x^2+y^2}{x^2y^2}}+\sqrt{\dfrac{2y^2+z^2}{y^2z^2}}+\sqrt{\dfrac{2z^2+x^2}{x^2z^2}}\)
\(P=\sqrt{\dfrac{2}{y^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{2}{z^2}+\dfrac{1}{y^2}}+\sqrt{\dfrac{2}{x^2}+\dfrac{1}{z^2}}\)
\(P\ge\sqrt{\left(\dfrac{\sqrt{2}}{x}+\dfrac{\sqrt{2}}{y}+\dfrac{\sqrt{2}}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=3\)
Đáp án của bạn ở đây: https://dethihsg.com/de-thi-hoc-sinh-gioi-toan-9-phong-gddt-cam-thuy-2011-2012/amp/
Rút gọn
P=\(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
b) tìm giá trị x để \(P\in Z\)
TA có :\(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}< \dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Mà ta lại có:\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}\)
Ta có \(0< \dfrac{1}{\sqrt{x}+1}< 1\)
\(\Rightarrow0< 1+\dfrac{1}{\sqrt{x}+1}< 2\)
\(\Rightarrow0< \dfrac{\sqrt{x}+1}{x+\sqrt{x}+2}< 2\)
Theo đề ra P đạt giá trị nguyên mà
0 < P <2
=> P=1
HAy\(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}=1\)
\(\Leftrightarrow\sqrt{x}+2=x+\sqrt{x}+1\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\) (ktmdk)
VẬy không giá trị nào của x đề P đạt giá trị nguyên.
a/ \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}=1+\dfrac{1}{2.2}+...+\dfrac{1}{n.n}\)
\(< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=1+1-\dfrac{1}{n}=2-\dfrac{1}{n}< 2\)
Lời giải:
\(S_n=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+..+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Để \(S_n\in\mathbb{Z}\Rightarrow \sqrt{n+1}-1\in\mathbb{Z}\Rightarrow \sqrt{n+1}\in\mathbb{Z}\)
Đặt \(\sqrt{n+1}=t\in\mathbb{N}>1\) do \(n>0\)
\(\Rightarrow n+1=t^2\Rightarrow t^2\leq 101\) do \(n\leq 100\)
\(\Rightarrow 0< t\leq \sqrt{101}\)
Mà \(t\in\mathbb{N}^*\Rightarrow t\in\left\{1;2;3;4;5;6;7;8;9;10\right\}\)
\(\Rightarrow n=t^2-1\in\left\{3; 8; 15; 24;35;48;63;80;99\right\}\)