Chứng minh : \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{2005\sqrt{2004}}< 2\)
Những câu hỏi liên quan
Chứng minh:
\(\dfrac{1}{1\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+....+\dfrac{1}{2004\sqrt{2005}}< 2\)
Chứng minh rằng : \(\dfrac{1}{2}\)+\(\dfrac{1}{3\sqrt{2}}\)+\(\dfrac{1}{4\sqrt{3}}\)+...+\(\dfrac{1}{2005\sqrt{2004}}\)
Chứng minh biểu thức đó <2
Với mọi \(n\in N^{\cdot}\), ta có
\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Leftrightarrow1< 2\left(n+1\right).\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
\(\Leftrightarrow0< n+1-2\sqrt{n+1}.\sqrt{n}+n\)
\(\Leftrightarrow0< \left(\sqrt{n+1}-\sqrt{n}\right)^2\)(Luôn đúng vì n thuộc N*)
Do đó: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...\dfrac{1}{2005\sqrt{2004}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2004}}-\dfrac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\dfrac{1}{\sqrt{2005}}\right)< 2\)
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CMR: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+....+\dfrac{1}{2005\sqrt{2004}}< 2\)
CMR: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+....+\dfrac{1}{2005\sqrt{2004}}< 2\)
Xét dạng tổng quát:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\)
\(< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2004}}\right)\)
\(< 2-\frac{2}{\sqrt{2004}}< 2\)
=>đpcm
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CMR 2004 < 1+ \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{1006009}}< 2005\)
Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{matrix}\right.\)
Thế vô giải tiếp
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chứng minh
\(\dfrac{3}{2}\)\(\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
rút gọn
D=\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\)\(-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
a)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
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b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)
\(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
D=2
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Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)
1) Chứng minh rằng: 1+dfrac{1}{2sqrt{2}}+dfrac{1}{3sqrt{3}}+...+dfrac{1}{nsqrt{n}} 2sqrt{2}left(nin Nright)
2) Chứng minh rằng: dfrac{2}{3}+sqrt{n+1} 1+sqrt{2}+sqrt{3}+...+sqrt{n} dfrac{2}{3}left(n+1right)sqrt{n}
3) 2sqrt{n}-3 dfrac{1}{sqrt{2}}+dfrac{1}{sqrt{3}}+...+dfrac{1}{sqrt{n}} 2sqrt{n}-2
4) dfrac{sqrt{2}-sqrt{1}}{2+1}+dfrac{sqrt{3}-sqrt{2}}{3+2}+...+dfrac{sqrt{n+1}-sqrt{n}}{n+1+n} dfrac{1}{2}left(1-dfrac{1}{sqrt{n+1}}right)
Đọc tiếp
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
