Chứng minh rằng: \(2\sqrt{n}+1+\sqrt{n}< \dfrac{1}{\sqrt{n}}\) với \(n\in\text{N*}\)
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1 tháng 1 2019
a)
+) Ta có: \(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\) \(=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\) (1)
+) Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\) \(=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-\left(n-1\right)}\)
\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)
Từ (1) và (2) ⇒ đpcm
Học toán vui vẻ! 
AH
Akai Haruma
Giáo viên
31 tháng 10 2019
Lời giải:
Liên hợp ta thấy:
\(2(\sqrt{n+1}-\sqrt{n})=2.\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(1)\)
\(2(\sqrt{n}-\sqrt{n-1})=2.\frac{n-(n-1)}{\sqrt{n}+\sqrt{n-1}}=\frac{2}{\sqrt{n}+\sqrt{n-1}}>\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{1}{\sqrt{n}}(2)\)
Từ \((1);(2)\Rightarrow 2(\sqrt{n+1}-\sqrt{n})< \frac{1}{\sqrt{n}}< 2(\sqrt{n}-\sqrt{n-1})\)
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Áp dụng vào bài toán:
\(S=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>1+2(\sqrt{3}-\sqrt{2})+2(\sqrt{4}-\sqrt{3})+...+2(\sqrt{101}-\sqrt{100})\)
\(\Leftrightarrow S>1+2(\sqrt{101}-\sqrt{2})>18(*)\)
Và:
\(S< 1+2(\sqrt{2}-\sqrt{1})+2(\sqrt{3}-\sqrt{2})+....+2(\sqrt{100}-\sqrt{99})\)
\(\Leftrightarrow S< 1+2(\sqrt{100}-\sqrt{1})=19(**)\)
Từ $(*); (**)$ suy ra $18< S< 19$ (đpcm)
24 tháng 6 2018
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
24 tháng 6 2018
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
28 tháng 5 2017
\(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}>\sqrt{n}\left(1\right)\)
Với \(n=2\), BĐT \(\left(1\right)\) trở thành \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}>\sqrt{2}\) (đúng)
Giả sử \(\left(1\right)\) đúng với \(n=k\), nghĩa là \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}>\sqrt{k}\left(2\right)\)
Ta chứng minh \(\left(1\right)\) đúng với \(n=k+1\). Thật vậy, từ \(\left(2\right)\) suy ra:
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}>\sqrt{k}+\dfrac{1}{\sqrt{k+1}}\)
Vì \(\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{\sqrt{k\left(k+1\right)}+1}{\sqrt{k+1}}>\sqrt{k+1}\)
Nên \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}>\sqrt{k+1}\)
Tức là \(\left(1\right)\) đúng với \(n=k+1\).
Theo nguyên lí quy nạp, (1) đúng với mọi số tự nhiên \(n>1\)
21 tháng 9 2021
Ta có: \(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}\)
\(=\dfrac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\)
\(\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\)
Ta lại có: \(\sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n-1}\right)}{\sqrt{n}+\sqrt{n-1}}\)
\(=\dfrac{n-n+1}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\)
\(\Rightarrow2\left(\sqrt{n}-\sqrt{n-1}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
21 tháng 9 2021
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(1\right)\)
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\RightarrowĐpcm\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 6 2021
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
AH
Akai Haruma
Giáo viên
12 tháng 1 2019
Lời giải:
Đặt \(P=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}\)
Ta có:
\(\frac{P}{2}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{4}}+...+\frac{1}{2\sqrt{n}}\)
\(< \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{n-1}+\sqrt{n}}(1)\)
Mà:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}{\sqrt{1}+\sqrt{2}}+\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{2}+\sqrt{3}}+\frac{(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})}{\sqrt{3}+\sqrt{4}}+....+\frac{(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})}{\sqrt{n-1}+\sqrt{n}}\)
\(=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{n}-\sqrt{n-1})\)
\(=\sqrt{n}-1(2)\)
Từ \((1);(2)\Rightarrow \frac{P}{2}< \sqrt{n}-1\Rightarrow P< 2\sqrt{n}-2\)
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Tương tự:
\(\frac{P}{2}>\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}+\frac{1}{2\sqrt{n}}=\sqrt{n}-\sqrt{2}+\frac{1}{2\sqrt{n}}\)
\(\Rightarrow P> 2\sqrt{n}-2\sqrt{2}+\frac{1}{\sqrt{n}}\)
Mà \(2\sqrt{n}-2\sqrt{2}+\frac{1}{\sqrt{n}}> 2\sqrt{n}-3\Rightarrow P>2\sqrt{n}-3\)
Ta có đpcm.
11 tháng 10 2021
\(a,\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\\ =3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\\ b,\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}\\ =\dfrac{\sqrt{n}-\sqrt{n+1}}{-1}=\sqrt{n+1}-\sqrt{n}\)
11 tháng 10 2021
a) \(\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\)
b) \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)