A= 2013/2014 + 2014/2015 + 2015/2013. So sánh A với 3
So sánh: 2013/2014+2014/2015+2015/2013 với 3
so sánh A=2013/2014 + 2014/2015 + 2015/2016 và B=2013+2014+2015/2014+2015+2016
A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)
\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)
\(Vậy:A>B\)
Đúng nha Nguyễn Bình Minh
so sánh:
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\) và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)
\(B=\frac{2013}{2014+2015+2016}+\frac{2014}{2014+2015+2016}+\frac{2015}{2014+2015+2016}\)
Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)
\(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)
\(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}>\frac{2013+2014+2015}{2014+2015+2016}\)
Vậy: \(A>B\)
So sánh các biểu thức sau: A =2013+2014/2014+2015 và B=2013/2014 + 2014/2015
A=\(\dfrac{2013+2014}{2014+2015}=\dfrac{2013}{2014+2015}+\dfrac{2014}{2014+2015}\)
B=\(\dfrac{2013}{2014}+\dfrac{2014}{2015}\)
Vì \(\dfrac{2013}{2014}>\dfrac{2013}{2014+2015}\); \(\dfrac{2014}{2015}>\dfrac{2014}{2014+2015}\) nên B>A
So sánh:
a) A=9^10 và B= ( 8^9+7^9+6^9+...+2^9+1^9)
b) P= 2013/2014 + 2014/2015 + 2015/2016 với Q= 2013+2014+2015 / 2014+2015+2016
Cho biểu thức A= 2013/2014 + 2014/2015 + 2015/2013. Hãy so sánh A với 3.( Mọi người giải đầy đủ giúp mình nhá! Cảm ơn!)
so sánh A và B
A=2013+2014 / 2014+2015
B=2013+2014 / 2014+2015
Cho A=2016/2015+2015/2014+2014/2013
So sánh A với 3
\(\frac{2016}{2015}>1;\frac{2015}{2014}>1;\frac{2014}{2013}>1.\)
\(\Rightarrow\frac{2016}{2015}+\frac{2015}{2014}+\frac{2014}{2013}>1+1+1=3\)
Vậy A>3
cho \(a=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\)
so sánh a với 3
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}=3,00000074\)
Vì 3,00000074 > 3 nên A > 3
Cho biểu thức A = \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\). Hãy so sánh A với 3.
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2013}{2013}+\frac{1}{2013}+\frac{1}{2013}=\left(\frac{2013}{2014}+\frac{1}{2013}\right)+\left(\frac{2014}{2015}+\frac{1}{2013}\right)+1\)
Ta có: \(\frac{2013}{2014}+\frac{1}{2013}>\frac{2013}{2014}+\frac{1}{2014}=\frac{2014}{2014}=1\)
\(\frac{2014}{2015}+\frac{1}{2013}>\frac{2014}{2015}+\frac{1}{2015}=\frac{2015}{2015}=1\)
=> A > 1+ 1 + 1 = 3