so sánh
\(\sqrt{8}+\sqrt{5}và\sqrt{7}+\sqrt{6}\)
\(\sqrt{2015}+\sqrt{2017}và\sqrt{2016}\)
giúp mình với mấy pạn 
THANKS
K
Khách
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Những câu hỏi liên quan
KY
19 tháng 9 2019
a) \(\sqrt{2017}-2\sqrt{2016}=\sqrt{2017}-\sqrt{8064}< 0< \sqrt{2016}\)
b) \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=8=\sqrt{64}>\sqrt{61}\)
c) \(\left(\sqrt{2016}+\sqrt{2014}\right)^2=4030+\sqrt{2014.2016}\)
\(\left(2\sqrt{2015}^2\right)=4030+\sqrt{2015.2015}\)
C/m được: \(\sqrt{2014.2016}< \sqrt{2015.2015}\)
\(\Rightarrow\left(\sqrt{2016}+\sqrt{2014}\right)^2< \left(2\sqrt{2015}\right)^2\)
\(\Rightarrow\sqrt{2014}+\sqrt{2016}< 2\sqrt{2015}\)
d) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=7=8-1=\sqrt{64}-1< \sqrt{65}-1\)
BC
3
OY
18 tháng 10 2021
Ta có: \(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015.2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016.2017}\)
\(2015.2018=2015.2017+2015=2017\left(2015+1\right)-2017+2015=2017.2016-2\)\(\Rightarrow2015.2018< 2016.2017\)
\(\Rightarrow4033+2\sqrt{2015.2018}< 4033+2\sqrt{2016.2017}\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\left(đpcm\right)\)
10 tháng 9 2023
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
5 tháng 10 2021
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 \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
ST
1
16 tháng 10 2017
Ta có \(\sqrt{2015}+\sqrt{2016}< \sqrt{2016}+\sqrt{2017}\)
mà \(\left(\sqrt{2015}-\sqrt{2016}\right)\cdot\left(\sqrt{2015}+\sqrt{2016}\right)\)\(=\left(\sqrt{2016}-\sqrt{2017}\right)\cdot\left(\sqrt{2016}+\sqrt{2017}\right)\)\(=1\)
Suy ra \(\sqrt{2015}-\sqrt{2016}>\sqrt{2016}-\sqrt{2017}\)
LT
3
N
1
binh phuong len
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40}\)
\(\left(\sqrt{7}+\sqrt{6}\right)^2=13+2\sqrt{42}\)
vi
\(2\sqrt{40}< 2\sqrt{42}\)
nen \(\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)