Chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt[3]{2}}+\frac{1}{4\sqrt[3]{3}}+...+\frac{1}{\left(n+1\right)\sqrt[3]{n}}< 3\)
Những câu hỏi liên quan
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
a, Chứng minh
\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, Áp dụng
\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}......+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
a)\(\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2.n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b)\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\( S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
\(a,\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}\left(n+1-n\right)}\)
\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{\sqrt{n-1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{400\sqrt{399}+399\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)
Xem thêm câu trả lời
Chứng minh:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\frac{1}{\sqrt{n+1}}\left(n\in N\right)\)
Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
.....
\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)
Chứng minh : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Ta có \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> biểu thức trên sẽ \(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+1}}< 2\)\(\left(dpcm\right)\)
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Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> Biểu thức trên sẽ bé hơn \(2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+1}}< 2\Rightarrowđpcm\)
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Chứng minh rằng\(\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< \sqrt{2}\)
Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Ta có : \(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\frac{1}{k\left(k+1\right)}\right)=\sqrt{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\sqrt{k}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+}}\right)\)
\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
Áp dụng : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+}}< 2\)
Vậy ta có điều phải chứng minh.
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1. Tính giá trị biểu thức: Asqrt{a^2+4ab^2+4b}-sqrt{4a^2-12ab^2+9b^4} với asqrt{2} ; b1
2. Đặt Msqrt{57+40sqrt{2}} ; Nsqrt{57-40sqrt{2}}. Tính giá trị của các biểu thức sau:
a) M-N
b) M^3-N^3
3. Chứng minh: left(frac{xsqrt{x}+3sqrt{3}}{x-sqrt{3x}+3}-2sqrt{x}right)left(frac{sqrt{x}+sqrt{3}}{3-x}right)1 (với xge0 và xne3)
4. Chứng minh: frac{left(sqrt{a}-sqrt{b}right)^2+4sqrt{ab}}{sqrt{a}+sqrt{b}}.frac{asqrt{b}-bsqrt{a}}{sqrt{ab}}a-b (a 0 ; b 0)
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Đọc tiếp
1. Tính giá trị biểu thức: \(A=\sqrt{a^2+4ab^2+4b}-\sqrt{4a^2-12ab^2+9b^4}\) với \(a=\sqrt{2}\) ; \(b=1\)
2. Đặt \(M=\sqrt{57+40\sqrt{2}}\) ; \(N=\sqrt{57-40\sqrt{2}}\). Tính giá trị của các biểu thức sau:
a) M-N
b) \(M^3-N^3\)
3. Chứng minh: \(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)=1\) (với \(x\ge0\) và \(x\ne3\))
4. Chứng minh: \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=a-b\) (a > 0 ; b > 0)
5. Chứng minh: \(\sqrt{9+4\sqrt{2}}=2\sqrt{2}+1\) ; \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}\) ; \(3-2\sqrt{2}=\left(1-\sqrt{2}\right)^2\)
6. Chứng minh: \(\left(\frac{1}{2\sqrt{2}-\sqrt{7}}-\left(3\sqrt{2}+\sqrt{17}\right)\right)^2=\left(\frac{1}{2\sqrt{2}-\sqrt{17}}-\left(2\sqrt{2}-\sqrt{17}\right)\right)^2\)
7. Chứng minh đẳng thức: \(\left(\frac{3\sqrt{2}-\sqrt{6}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}=-\frac{4}{3}\)
8.Chứng minh: \(\frac{2002}{\sqrt{2003}}+\frac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
9. Chứng minh rằng: \(\sqrt{2000}-2\sqrt{2001}+\sqrt{2002}< 0\)
10. \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) ; \(\frac{7}{5}< \frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}< \frac{29}{30}\)
1) Chứng minh: \(2\sqrt{n}-3< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}-2\forall n\ge2\)
2) Thu gọn: \(A=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)