trên mạng có đó Triphai Tyte

cho mình xin link đi

a) Có \(\sqrt{25}=5;\sqrt{45}< \sqrt{49}=7\)

\(\Rightarrow\sqrt{25}+\sqrt{45}< 5+7=12\)

Vậy \(\sqrt{25}+\sqrt{45}< 12.\)

b) có \(\left(\sqrt{2013}+\sqrt{2015}\right)^2=2013+2015+2\sqrt{2013}.\sqrt{2015}\)\(=4028+2\sqrt{2013.2015}\)

\(\left(2\sqrt{2014}\right)^2=4.2014=4028+2.2014=4028+2\sqrt{2014^2}\)

Xét \(2014^2-2013.2015=2014.\left(2013+1\right)-2013\left(2014+1\right)\)

\(=2013.2014+2014-2013.2014-2013=1>0\)

\(\Rightarrow2\sqrt{2013.2015}< 2\sqrt{2014^2}\)

Hay \(\left(\sqrt{2013}+\sqrt{2015}\right)^2< \left(2\sqrt{2014}\right)^2\)

\(\Rightarrow\sqrt{2013}+\sqrt{2015}< 2\sqrt{2014}\)
Vậy \(\sqrt{2013}+\sqrt{2015}< 2\sqrt{2014}.\)

c) Có \(\left(\sqrt{2014}-\sqrt{2013}\right)\left(\sqrt{2014}+\sqrt{2013}\right)=2014-2013=1\)\(\rightarrow\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)

Mà \(\sqrt{2014}>\sqrt{2013};\sqrt{2013}>\sqrt{2012}\)

\(\rightarrow\sqrt{2014}+\sqrt{2013}>\sqrt{2013}+\sqrt{2012}\)

Hay \(\dfrac{1}{\sqrt{2014}+\sqrt{2013}}< \dfrac{1}{\sqrt{2013}+\sqrt{2012}}\)

Tương tự, ta có \(\dfrac{1}{\sqrt{2013}+\sqrt{2012}}=\sqrt{2013}-\sqrt{2012}\)

\(\Rightarrow\sqrt{2014}-\sqrt{2013}< \sqrt{2013}-\sqrt{2012}\)

Vậy \(\sqrt{2014}-\sqrt{2013}< \sqrt{2013}-\sqrt{2012}.\)

lop8. thi ap bdt nhu thanh song,

a)

VT=√25+√45<√2(25+45)=√140<√144=12=VP

b)

VT=√2013+√2015<√[2(2013+2015)]=√[4.2014]=2√(2014)=VP.

c) C=VT-VP

√2014+√2012-2√2012

kq(b)=> C<0

VT<VP

B=\(2\sqrt{2012}\)nhé

nhanh giúp mình với

√2012+ √2014 và  2√2013

\(\sqrt{2012}+\sqrt{2014}< 2\sqrt{2013}\)

√2012 + √2014 < 2√2013

\(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015\cdot2018}\)

\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016\cdot2017}\)

\(2015\cdot2018=2015\cdot2017+2015=2017\cdot\left(2015+1\right)-2017+2015\)

\(=2017\cdot2016-2\)

\(\Rightarrow2015\cdot2018< 2016\cdot2017\)

\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\)

có bạn nào giải thích cho mình từ đoạn 2015.2018=2015.2017+2015 trở đi được k? mình cảm ơn