2016*2016...2015*2017
Những câu hỏi liên quan
(1/2+2015/2016+2016/2017+1)(2015/2016+2016/2017+7/22)-(1/2+2015/2016+2016/2017)(2015/2016+2016/2017+7/22+1)
Tính
(1/2 + 2015/2016 + 2016/2017 + 1)(2015/2016 + 2016/2017 + 7/22) - (1/2 + 2015/2016 + 2016/2017)(2015/2016 + 2016/2017 + 7/22 + 1)
Giúp mk với
(1/2+2015/2016+2016/2017+1)(2015/2016+2016/2017+7/22)-(1/2+2015/2016+2016/2017)(2015/2016+2016/2017+7/22+1)
(1+2015/2016+2016/2017+1/2).(2015/2016+2016/2017+7/22)-(2015/2016+2016/2017+1/2).(2015/2016+2016/2017+7/22+1)
tính tổng trên
( trình bày cách tính
(1/2+2015/2016+216/2017+1)(2015/2016+2016/2017+7/22)-(1/2+2015+2016)(2015/2016+2016/2017+7/22+1)
A = 2015/2016 + 2016/2017 + 2017/2018 và B = (2015 + 2016 + 2017)/(2016 + 2017 + 2018)
A=[(-2015)^2016.(-2016^2017)+(-2016)^2017.(-2015^2016)].(-2017)^2018 tính biểu thức A
Ta có : \(A=\left(\left(-2015\right)^{2016}.-2016^{2017}+\left(-2016\right)^{2017}.-2015^{2016}\right).\left(-2017\right)^{2018}\)
\(=\left(2015^{2016}.-2016^{2017}-2016^{2017}.-2015^{2016}\right).2017^{2018}\)
\(=\left(2015^{2016}-2015^{2016}\right).2017^{2018}.\left(-2016^{2017}\right)\)
\(=0.2017^{2018}.\left(-2016^{2017}\right)=0\)
Đúng 2
Bình luận (0)
Giải:
\(A=\left[\left(-2015\right)^{2016}.\left(-2016^{2017}\right)+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}.\left(-2016\right)^{2017}+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}+\left(-2015^{2016}\right)\right].\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0.\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0\)
Đúng 0
Bình luận (0)
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)
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 \(A>B\)
So sánh A và B :
A = 2015 . 2016 + 2017 / 2015 . 2016 + 2016
B = 2015 . 2016 + 2018 / 2015 . 2016 + 2017
Đặt 2015.2016+2016=n
suy ra A=(n+1)/n và B=(n+2)/(n+1)
Ta có A - B=(n+1)/n -(n+2)/(n+1)=((n+1)2-n(n+2))/n(n+1)=(n2+2n+1-n2-2n)/n(n+1)=1/n(n+1)
Vì A-B lớn hơn 0 nên A>B
Đúng 0
Bình luận (0)