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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

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n ejmfjnhcy

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 \(\sqrt{2015}-\sqrt{2014}< \sqrt{2014}-\sqrt{2013}\)

\(a\))Ta có:\(2\sqrt{3}=\sqrt{12}\)

             \(3\sqrt{2}=\sqrt{18}\)

Vì \(\sqrt{12}< \sqrt{18}\)

⇒\(2\sqrt{3}< 3\sqrt{2}\)

\(b\))Ta có:\(2\sqrt{3}+1=\sqrt{12}+1\)

             \(4=3+1=\sqrt{9}+1\)

Vì \(\sqrt{12}+1>\sqrt{9}+1\)

⇒\(2\sqrt{3}+1>4\)

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

\(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}-\frac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

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