so sánh A=\(\frac{2019}{\sqrt{2020}}+\frac{2020}{\sqrt{2019}}\)
và B =\(\sqrt{2019}+\sqrt{2020}\)
K
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DT
17 tháng 7 2019
1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)
\(\Rightarrow1+2019^2=2020^2-2.2019\)
\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)
\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)
\(=2020\)
Vậy M=2020.
2) Xét : \(k\in N;k\ge2\)ta có:
\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)
\(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)
\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)
Cho \(k=3,4,...,2020.\)Ta có:
\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)
Vậy \(N=2018\frac{1009}{2020}.\)
AH
Akai Haruma
Giáo viên
14 tháng 8 2019
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\)
\(=1-\frac{1}{\sqrt{2020}}\)
NC
28 tháng 9 2020
Xét phân thức phụ sau, với n nguyên dương lớn hơn 1 ta có:
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}\right)^2\sqrt{n}}=2\left(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)\sqrt{n}}\right)\)
\(=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán ta được:
\(A=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\right)\)
\(A=2-\frac{2}{\sqrt{2020}}< 2=B\)
Vậy A < B
KS
1
21 tháng 10 2020
Ta có: \(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
\(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}+\sqrt{2019}\right)\left(\sqrt{2020}-\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
Do \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2020}+\sqrt{2019}}\) => \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)
19 tháng 10 2020
\(^6\sqrt{2019} = b, ^6\sqrt{2020} = a \\ Then, A = a^3 - b^3; B = a^2 -b^2\\ \Rightarrow A > B \)
Lời giải:
Đặt \(\sqrt{2019}=a; \sqrt{2020}=b\) $(a,b>0)$
Ta có:
\(A-B=\frac{a^2}{b}+\frac{b^2}{a}-a-b\)
\(=(\frac{a^2}{b}-b)+(\frac{b^2}{a}-a)=\frac{a^2-b^2}{b}-\frac{a^2-b^2}{a}=(a^2-b^2)(\frac{1}{b}-\frac{1}{a})=\frac{(a-b)^2(a+b)}{ab}>0\) với mọi $a\neq b; a,b>0$
Do đó A>B$