So sánh giá trị của A và B với:
A = \(\dfrac{2017^{2016-1}}{2017^{2017-1}}\) B = \(\dfrac{2017^{2015+1}}{2017^{2016+1}}\)
So sánh \(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\) và \(B=2018\)
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
So sánh :
A = \(\dfrac{2017^{2015}+1}{2017^{2016}+1}\)
B = \(\dfrac{2017^{2014}+1}{2017^{2015}+1}\)
Ta có :
\(2017A=\dfrac{2017\left(2017^{2015}+1\right)}{2017^{2016}+1}\)
\(=\dfrac{2017^{2016}+2017}{2017^{2016}+1}\)
\(=\dfrac{\left(2017^{2016}+1\right)+2016}{2017^{2016}+1}\)
\(=\dfrac{2017^{2016}+1}{2017^{2016}+1}\) + \(\dfrac{2016}{2017^{2016}+1}\)
\(=1+\dfrac{2016}{2017^{2016}+1}\) (1)
Tương tự :
\(2017B=\dfrac{2017\left(2017^{2014}+1\right)}{2017^{2015}+1}\)
\(=\dfrac{2017^{2015}+2017}{2017^{2015}+1}\)
\(=1+\dfrac{2016}{2017^{2016}+1}\) (2)
Từ (1) và (2) => \(2017A< 2017B\)
=> \(A< B\)
So sánh giá trị của A và B với:
A = \(\frac{2017^{2016-1}}{2017^{2017-1}}\) and B = \(\frac{2017^{2015+1}}{2017^{2016+1}}\)
Arigato so much!
\(A=\frac{2017^{2016-1}}{2017^{2017-1}}=\frac{2017^{2015}}{2017^{2016}}=\frac{2017^{2015}}{2017^{2015}.2017}=\frac{1}{2017}\)(1)
\(B=\frac{2017^{2015+1}}{2017^{2016+1}}=\frac{2017^{2016}}{2017^{2017}}=\frac{2017^{2016}}{2017^{2016}.2017}=\frac{1}{2017}\)(2)
Từ (1) và (2)\(\Rightarrow\)A = B
Homie ơi, giúp mình với:
1) So sánh:
a) -2016/2017 và -2015/2016
b) 2017/-2016 và 2016/-2015
a)\(\frac{2016}{2017}< 1;\frac{2015}{2016}< 1\)
b)\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và
\(\frac{2016}{2017}< 1;\frac{2016}{2015}< 1\)
\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và \(\frac{2015}{2016}\)< \(\frac{2017}{2016}\)và \(\frac{2016}{2015}\)
so sánh 2016^2016 +1 / 2016^2015+1 và 2017^2017 +1 / 2017^2016 +1
so sánh : A= 2015/2016 + 2016/2017 + 2017/2018 và B= 2015+2016+2017 / 2016+2017+2018
A=2015/2016+2016/2017+2017/2018>2015/2018+2016/2018+2017/2018
=6048/2018>1
B=2015+2016+2017/2016+2017+2018=6048/6051<1
=>A>B
so sánh : A= 2015/2016 + 2016/2017 + 2017/2018 và B= 2015+2016+2017 / 2016+2017+2018
Có: B = 2015 + 2016 + 2017/2016 + 2017 + 2018
B= 2015 / (2015 + 2016+2017) + 2016/(2016+2017+2018) + 2017/(2016 + 2017 + 2018)
vì 2015/2016 > 2015/(2016 + 2017+2018) ; 2016/2017>2016/(2016+2017+2018) ; 2017/2018 > 2017/(2016+2017+2018)
=> A>B
có ai là ARMY ko nếu là ARMY thì mọi người cày view chưa
so sánh : A= 2015/2016 + 2016/2017 + 2017/2018 và B= 2015+2016+2017 / 2016+2017+2018
So sánh(không dùng bảng số hay máy tính cầm tay)
a)\(\dfrac{1}{7}\sqrt{51}\) với \(\dfrac{1}{9}\sqrt{150}\)
b)\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)