phép đầu là \(\frac{2017.2018+2019}{2018.2019-2017}\)
phép đầu là: \(\frac{2017.2018+2019}{2018.2019-2017}\)
phép đầu là \(\frac{2017.2018+2019}{2018.2019-2017}\)
phép đầu là: \(\frac{2017.2018+2019}{2018.2019-2017}\)
So sánh A và B biết :
A = \(-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}\)
B = \(-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\)
Giúp mk nha mn
Cho \(A=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\). Chứng tỏ A ko phải là 1 số nguyên.
Mk cần gấp. Mai nộp rồi!!!
so sánh a và b biết a=2016/2017+2017/2018+2018/2019+2019/2016 và b=1/8+1/9+1/10+...+1/63
Tìm x khi biết x:☺
\(\left(\frac{x-4}{2017}\right)+\left(\frac{x-3}{2018}\right)+\left(\frac{x-2}{2019}\right)+\left(\frac{x-1}{2020}\right)=4\)
gip mik với mik cần gấppppppppppppppppp
So sánh \(A=\frac{2^{2019}}{2^{2019}+3^{2019}}+\frac{3^{2019}}{3^{2019}+5^{2019}}+\frac{5^{2019}}{5^{2019}+2^{2019}}\)với \(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
a)\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
b)\(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)
c)\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\)
d)\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
ai xong nhanh nhất và đúng em xin gửi 2 SP ạ
\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2019}}+\frac{1}{3^{2020}}< \frac{1}{2}\)
Tính B=\(2020.\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).....\left(\frac{1}{2019}-1\right)\left(\frac{1}{2020}-1\right)\)
B1 : Tìm GTNN :
\(\left(x+2020\right)^4+\left|y-2019\right|-2018\)
B2 : Tính :
\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2019}.\left(1+2+3+...+2019\right)\)