Tính S=\(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+....+\frac{2020}{2020^4+2020^2+2020}\)
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 7 2020
1. Kẻ \(BH\perp AC\Rightarrow BH=AB.sin60^0=2\sqrt{2}.\frac{\sqrt{3}}{2}=\sqrt{6}\)
\(\Rightarrow S_{ABC}=\frac{1}{2}BH.AC=3\sqrt{2}\)
2. \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right)\left(\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n\left(n+1\right)^2}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
\(S=2020\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\right)\)
\(=2020\left(1-\frac{1}{\sqrt{2020}}\right)=2020-\sqrt{2020}\)
21 tháng 10 2020
Đk: \(\forall x\in R\)
Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=2021\)
Lập bảng xét dầu
x -2 1
x - 1 - | - 0 +
x + 2 - 0 + | -
Xét các TH xảy ra :
TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021
<=> -2x = 2022 <=> x = -1011 (tm)
TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021
<=> 0x = 2018 (vô lí) => pt vô nghiệm
TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021
<=> 2x = 2020 <=> x = 1010 (tm)
Vậy S = {-1011; 1010}
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 \(N=2018\frac{1009}{2020}.\)
29 tháng 3 2020
Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)
6 tháng 6 2019
@Thế Vĩ@
\(P=\sqrt{2}.\frac{\sqrt{2020}-\sqrt{2}}{2}=\sqrt{2}.\frac{\sqrt{2}\left(\sqrt{1010}-1\right)}{2}=2.\frac{\sqrt{1010}-1}{2}=\sqrt{1010}-1\)
22 tháng 8 2020
TA XÉT PHÂN THỨC TỔNG QUÁT SAU:
\(A=\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}\)
\(A=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(A=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(A=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
THAY LẦN LƯỢT CÁC GIÁ TRỊ n từ 1 => 2021 vào ta được:
=> \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
=> \(A=1-\frac{1}{\sqrt{2021}}=\frac{\sqrt{2021}-1}{\sqrt{2021}}\)
VẬY \(A=\frac{\sqrt{2021}-1}{\sqrt{2021}}.\)
NN
22 tháng 8 2020
Ta có: \(\frac{1}{\left(a-1\right)\sqrt{a}+a.\sqrt{a-1}}=\frac{a-\left(a-1\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}=\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}\)
\(=\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a-1}}-\frac{\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}=\frac{1}{\sqrt{a-1}}-\frac{1}{\sqrt{a}}\)
Thay lần lượt các giá trị của a bằng \(2;3;4;........;2021\)ta được:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2021}}=1-\frac{1}{\sqrt{2021}}\)