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GD
1
XO
9 tháng 3 2020
Ta có A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\)(1)
=> 3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)(2)
Lấy (2) trừ (1) theo vế ta có :
3A - A = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\right)\)
2A = \(1-\frac{1}{3^{2019}}\)
Khi đó : \(\left(2A+\frac{1}{3^{2019}}\right).x=2\)
\(\Leftrightarrow\left(1-\frac{1}{3^{2019}}+\frac{1}{3^{2019}}\right).x=2\)
\(\Rightarrow x=2\)
HP
0
BL
0
BL
1
B
8 tháng 2 2020
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
---------------------
Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
9 tháng 10 2019
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:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
NB
19 tháng 12 2019
B1:
\(A=\left(x+2020\right)^4+\left|y-2019\right|-2018\)
+Có: \(\left(x+2020\right)^4\ge0với\forall x\\\left|y-2019\right|\ge0với\forall y\\\Rightarrow \left(x+2020\right)^4+\left|y-2019\right|-2018\ge-2018\\ \Leftrightarrow A\ge-2018 \)
+Dấu "=" xảy ra khi
\(\left(x+2020\right)^4=0\\ \Leftrightarrow x=-2020\)
\(\left|y-2019\right|=0\\ \Leftrightarrow y=2019\)
+Vậy \(A_{min}=-2018\) khi \(x=-2020,y=2019\)
\(1+\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+...+\frac{1}{\left(2019+1\right).2019:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2\left(1-\frac{1}{2020}\right)=\frac{2019}{1010}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2019}\)
\(=1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2039190}\)
\(=2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4078380}\right)\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2.\left(1-\frac{1}{2020}\right)\)
\(=2.\frac{2019}{2020}\)
\(=\frac{2019}{1010}\)