\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{2020}\left(1+2+3+...+2020\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2020}.\frac{2020.2021}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)

\(=\frac{\left[\left(2021-2\right)+1\right]\left(2021+2\right)}{2}:2\)

\(=1021615\)

Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)

\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)

\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)

a)

\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)

      =\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)

      \(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)

Vay P=a+1

phan b,c ap dung phan a la ra

CM bài toán phụ: \(x+y+z=0\) 

CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương

Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)

\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(Q=2021-\frac{1}{2021}=...\)

Phần b mượn bài Upin ta có:

\(P=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

\(P=2020+1\)

\(P=2021\)

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

Ta có:

\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)

\(\Rightarrow A=16+\frac{\frac{2}{9}+\frac{2}{10}+\frac{2}{11}+...+\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)

\(\Rightarrow A=16+\frac{2\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)

\(\Rightarrow A=16+\frac{2}{\frac{1}{3}}\)

\(\Rightarrow A=16+\left(2:\frac{1}{3}\right)\)

\(\Rightarrow A=16+\left(2.3\right)\)

\(\Rightarrow A=16+6\)

\(\Rightarrow A=22\)

              Vậy\(A=22\)

A = 16 + (2/9+2/10+....+2/2020)/(1/27+1/30+.....+1/6060)

   = 16 + 6

   = 22

Tk mk nha

c) Áp dụng công thức \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\),ta được:

\(Q=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(=1+1+1+...+1-\frac{1}{2021}\)

\(=2021-\frac{1}{2021}=\frac{4084440}{2021}\)

\(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge2\sqrt{\frac{a^2}{a^2}}+2\sqrt{\frac{b^2}{b^2}}+2\sqrt{\frac{c^2}{c^2}}=6\)

Dấu = xảy ra khi a^4=b^4=c^4=1 <=> \(a=\pm1;b=\pm1;c\pm1\)

-> B = 3

Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)

Được cập nhật 41 giây trước (17:23)

Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020 
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5 
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1

là luc sắp chết mày yêu ai?