So sánh
C=\(\sqrt{2015}-\sqrt{2013}\)
D=\(\sqrt{2013}-\sqrt{2012}\)
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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
\(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015\cdot2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016\cdot2017}\)
\(2015\cdot2018=2015\cdot2017+2015=2017\cdot\left(2015+1\right)-2017+2015\)
\(=2017\cdot2016-2\)
\(\Rightarrow2015\cdot2018< 2016\cdot2017\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\)
có bạn nào giải thích cho mình từ đoạn 2015.2018=2015.2017+2015 trở đi được k? mình cảm ơn