2018-2017+2016-2015+.....+2-1
Những câu hỏi liên quan
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
so sánh 2 p/s A=2015/2016+2016/2017+2017/2018 va B=2015+2016+2017/2016+2017+2018
Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
Tính:
A=2019/2018 - 2018/2017 + 2017/2016 - 2016/2015
B=1/2019 - 1/2018 + 1/2017 - 1/2016
C=1/2017 - 1/2016 + 1/2015 - 1/2014
A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
a)
Ta có: 2015/2016=1-1/2016
2016/2017=1--1/2020.So sánh 1/2016 và 1/2017 được 1/2016>1/2017
Suy ra 2015/2016<2016/2017
b) 2018/2018=1
2019/2018>1
Vậy 2018/2018 <2019/2018
CHÚC BẠN HỌC TỐT NHÉ!!!
A = 2015/2016 + 2016/2017 + 2017/2018 và B = (2015 + 2016 + 2017)/(2016 + 2017 + 2018)
( 2016 * 2017 + 2018 * 2 + 2015 ) : [ ( 2018 : 2017 - 2017 * 2015 ) + 2016 ]
2018 * 20182017 - 2017 * 20172018
2018 * 20182017 - 2017 * 20182018
Bai nay la bai tinh nhanh nha
Khó quá zậy
Nhưng mình bít kết quả rồi
đó là : -1,002482622
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
tính nhanh:
a) (2016*2017+2018*2+2015):[(2018*2017-2017*2015)+2016]
b)2018*20182017-2017*20182018
c)1+2-3-4+5+6-7-8+...+298-299-300+301+302
so sánh
P=2015/2016+2016/2017+2017/2018 và Q=2015+2016+2017/2016+2017+2018
Ta có:\(Q=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì \(\hept{\begin{cases}\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\\\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\\\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\end{cases}}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow P>Q\)
Vậy P > Q
Đúng 0
Bình luận (0)