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Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014
a. Ta có \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\to\frac{1}{\sqrt{2016}+\sqrt{2015}}
a) Ta có: \(2\sqrt{3}=\sqrt{4\cdot3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{9\cdot2}=\sqrt{18}\)
mà \(\sqrt{12}< \sqrt{18}\)(vì 12<18)
nên \(2\sqrt{3}< 3\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{3}+1\right)^2=8+4\sqrt{3}+1=9+4\sqrt{3}\)
\(4^2=16=9+7\)
mà \(4\sqrt{3}< 7\left(\sqrt{48}< \sqrt{49}\right)\)
nên \(\left(2\sqrt{3}+1\right)^2< 4^2\)
hay \(2\sqrt{3}+1< 4\)
c) Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)
\(\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)
Ta có: \(\sqrt{2015}+\sqrt{2014}>\sqrt{2013}+\sqrt{2014}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2015}+\sqrt{2014}}< \dfrac{1}{\sqrt{2013}+\sqrt{2014}}\)
hay \(\sqrt{2015}-\sqrt{2014}< \sqrt{2014}-\sqrt{2013}\)
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
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}\le\sqrt{\frac{a+b}{2}}\) :
Xét : \(N-M=2\sqrt{2014}-\left(\sqrt{2015}+\sqrt{2013}\right)\)
Theo bđt trên thì \(\frac{\sqrt{2013}+\sqrt{2015}}{2}\le\sqrt{\frac{2013+2015}{2}}\Leftrightarrow\sqrt{2013}+\sqrt{2015}\le2\sqrt{2014}\)
\(\Rightarrow N-M>0\Rightarrow N>M\)
1) Ta có bđt sau : \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m)
Áp dụng : \(\frac{\sqrt{2005}+\sqrt{2007}}{2}< \sqrt{\frac{2005+2007}{2}}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
2) Xét : \(A-B=2\sqrt{2014}-\left(\sqrt{2013}+\sqrt{2015}\right)\)
Theo câu 1) , ta dễ dàng c/m được \(2\sqrt{2014}>\sqrt{2013}+\sqrt{2015}\)
Do đó A - B > 0 => A > B
2) Bình phương 2 vế ta có:
\(A^2=2014-2013=1\)
\(B^2=2015-2014=1\)
=>A=B