So sánh \(\frac{a}{b}\)và \(\frac{a+2015}{b+2015}\)
so sánh:
a) A=\(\frac{2015^{2016}+1}{2015^{2017}+1}\)và B=\(\frac{2015^{2015}+1}{2015^{2016}+1}\)
Vì \(2015^{2016}+1< 2015^{2017}+1\Rightarrow\frac{2015^{2016}+1}{2015^{2017}+1}< 1\)
\(\Rightarrow A=\frac{2015^{2016}+1}{2015^{2017}+1}< \frac{2015^{2016}+1+2014}{2015^{2017}+1+2014}=\frac{2015\left(2015^{2015}+1\right)}{2015\left(2015^{2016}+1\right)}=\frac{2015^{2015}+1}{2015^{2016}+1}=B\)
Vậy \(A< B\)
\(2015A=\frac{2015^{2017}+2015}{2015^{2017}+1}=\frac{2015^{2017}+1+2014}{2015^{2017}+1}=1+\frac{2014}{2015^{2017}+1}\)
\(2015B=\frac{2015^{2016}+2015}{2015^{2016}+1}=\frac{2015^{2016}+1+2014}{2015^{2016}+1}=1+\frac{2014}{2015^{2016}+1}\)
vì \(\frac{2014}{2015^{2017}+1}< \frac{2014}{2015^{2016}+1}\)
nên \(2015A< 2015B\)
=> \(B>A\)
Ta có: 2015^2016+1<2015^2017 +1
=> 2015^2016 +1/ 2015^2017+1 <1
=> A= 2015^2016 +1/ 2015^2017+1 < 2015^2016+1+2014/2015^2017+1+2014=2015^2015+1/2015^2016+1=B
Vậy A<B
---k mk bn nhé--- Lâu rồi mới lên :*
So sánh : \(A=\frac{2015^{2016}+1}{2015^{2015}+1}\) và \(B=\frac{2014^{2015}+1}{2014^{2014}+1}\)
A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)
B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)
Rồi bạn tự so sánh nha
\(B=\frac{10^{2015}}{10^{2015}-3}\)\(A=\frac{10^{2015}+2}{10^{2015}-1}\).So sánh A và B.
so sánh A=\(\frac{2015^{2016}+1}{2015^{2017}+1}\) và B=\(\frac{2015^{2017}+1}{2015^{2018}+1}\)
Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)ta có:
\(B=\frac{2015^{2017}+1}{2015^{2018}+1}< \frac{2015^{2017}+1+2014}{2015^{2018}+1+2014}=\frac{2015^{2017}+2015}{2015^{2018}+2015}\)
\(=\frac{2015\left(2015^{2016}+1\right)}{2015\left(2015^{2017}+1\right)}=\frac{2015^{2016}+1}{2015^{2017}+1}\)
\(\Rightarrow\frac{2015^{2017}+1}{2015^{2018}+1}< \frac{2015^{2016}+1}{2015^{2017}+1}\)
Vậy \(B< A\)
Hay \(A>B\)
So sánh: A=\(\frac{20^{2015}+1}{20^{2015}-1}\)Và \(B=\frac{20^{2015}-1}{20^{2015}-3}\)
Ta có
\(A=\frac{20^{2015}+1}{20^{2015}-1}=\frac{20^{2015}-1+2}{20^{2015}-1}=1+\frac{2}{20^{2015}-1}\)
\(B=\frac{20^{2015}-1}{20^{2015}-3}=\frac{20^{2015}-3+2}{20^{2015}-3}=1+\frac{2}{20^{2015}-3}\)
Vì \(1+\frac{2}{20^{2015}-1}< 1+\frac{2}{20^{2015}-3}\)
\(\Rightarrow A< B\)
\(A=\frac{20^{2015}+1}{20^{2015}-1}=\frac{20^{2015}-1+2}{20^{2015}-1}=1+\frac{2}{20^{2015}-1}\)
\(\frac{20^{2015}-1}{20^{2015}-3}=\)\(\frac{20^{2015}-3+2}{20^{2015}-3}=1+\frac{2}{20^{2015}-3}\)
vì \(1+\frac{2}{20^{2015}-1}>1+\frac{2}{20^{2015}-3}\)
vì tử của hai phân số này bằng nhau nên chúng ta so sánh mẫu, mẫu nào lớn hơn thì mẫu ấy bé hơn nhé
\(\Rightarrow A< B\)
So sánh A=\(\frac{2015}{-2014}\) và B=\(\frac{-2016}{2015}\) ta được A.......B
a) So sánh \(\frac{2013}{2015}\) và \(\frac{2014}{2016}\)
b) So sánh \(\frac{2013+2014}{2014+2015}\) và \(\frac{2013}{2014}+\frac{2014}{2015}\)
a)\(\frac{2013}{2015}< \frac{2014}{2016}\)
b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)
ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........
So sánh:
A=\(\frac{2015}{\sqrt{2016}}+\frac{2016}{\sqrt{2015}}\) và B=\(\sqrt{2015}+\sqrt{2016}\)
Có: \(\sqrt{2015}< \sqrt{2016}\)
=>\(\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2016}}\)
=>\(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>0\)
=>\(\sqrt{2015}+\sqrt{2016}+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
=>\(\left(\sqrt{2015}+\frac{1}{\sqrt{2015}}\right)+\left(\sqrt{2016}-\frac{1}{\sqrt{2016}}\right)>\sqrt{2015}+\sqrt{2016}\)
=>\(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
3, So sánh
\(A=\frac{2015^{2015}+1}{2015^{2016}+1}\)và \(B=\frac{2015^{2015}+1}{2015^{3016}+1}\)
nhìn cái đề là thấy A và B cùng tử
mẫu của A < mẫu của B thì
A>B
Vì mẫu lớn hơn thì nó nhỏ hơn
ma tử bằng nhau
=> A >B
So sánh:
A=\(\frac{2015}{\sqrt{2016}}+\frac{2016}{\sqrt{2015}}\) và B=\(\sqrt{2015}+\sqrt{2016}\)