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Đặt \(A=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{225}}\)
\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+...+\dfrac{2}{\sqrt{225}+\sqrt{225}}\)
\(\Rightarrow A< \dfrac{2}{\sqrt{2}+\sqrt{1}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}+...+\dfrac{2}{\sqrt{225}+\sqrt{224}}=\)
\(=2[\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+(\sqrt{225}-\sqrt{224})]\)
\(\Leftrightarrow A< 2.\left(\sqrt{225}-1\right)=2.14=28\left(đpcm\right)\)
Bài toán tổng quát:Chứng minh BĐT sau với \(n\in N;n\ge2\)
\(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
Đặt B = \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{50}}\)
= \(1+2\left(\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{2\sqrt{50}}\right)\)
Đặt \(A=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{2\sqrt{50}}\)
Xét A < \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{49}+\sqrt{50}}\)
=> A < \(\dfrac{\sqrt{2}-\sqrt{1}}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+...+\dfrac{\sqrt{50}-\sqrt{40}}{1}\)
=> A < -1 + \(\sqrt{50}\)
=> 2A < -2 + \(10\sqrt{2}\)
=> 2A + 1 = B < -2 + \(10\sqrt{2}\) + 1
=> B < -1 + \(10\sqrt{2}\) < \(10\sqrt{2}\) (1)
Xét \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
=> \(\dfrac{1}{\sqrt{1}}>2\left(\sqrt{2}-\sqrt{1}\right)\)
\(\dfrac{1}{\sqrt{2}}>2\left(\sqrt{3}-\sqrt{2}\right)\)
\(\dfrac{1}{\sqrt{3}}>2\left(\sqrt{4}-\sqrt{3}\right)\)
...
\(\dfrac{1}{\sqrt{50}}>2\left(\sqrt{51}-\sqrt{50}\right)\)
=> B > 2(\(\sqrt{51}-\sqrt{1}\))
=> B >-2 + \(10\sqrt{2}\) > \(5\sqrt{2}\)
Cảm ơn bạn nha. Mà bạn bị nhầm 49 thành 40 ở dòng thứ 5 đó.
Lời giải:
Gọi biểu thức vế trái là $A$. Ta có:
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{100}}\)
\(< \frac{1}{2}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
Mà, thực hiện liên hợp để trục căn thức dưới mẫu thì ta có:
\(\frac{1}{2}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}=\frac{1}{2}+\frac{\sqrt{2}-\sqrt{1}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+..+\frac{\sqrt{100}-\sqrt{99}}{1}\)
\(=\frac{1}{2}+\sqrt{100}-1=\frac{19}{2}\)
Do đó:
\(\frac{A}{2}< \frac{19}{2}\Rightarrow A< 19< 20\) (đpcm)
\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
ĐK : a >= 0 , a khác 1
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)
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 \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
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 \(\left(4\right)\Rightarrow18< S< 20\)
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! 