Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

Thầy phynit, cô @Cẩm Vân Nguyễn Thị, các bạn hok giỏi Toán: @Nguyễn Huy Tú, @Nguyễn Trần Thành Đạt, ..................

Giups em vs

tớ biết làm bài này

Hình như cậu ko cân mk

A=\(\dfrac{2014^{2013}+1}{2014^{2014}+1}\)<\(\dfrac{2014^{2013}+1+2013}{2014^{2014}+1+2013}\)

=\(\dfrac{2014^{2013}+2014}{2014^{2014}+2014}\)=\(\dfrac{2014\left(2014^{2012}+1\right)}{2014\left(2014^{2013}+1\right)}\)

=\(\dfrac{2014^{2012}+1}{2014^{2013}+1}\)=B

Vậy A<B

khó quá

Không ai trả lời đc đành phải tự trả lời thôi :))

A=1/2! - 2/3! - 3/4! - .... - 2013/2014!

=1/2! - (2/3! + 3/4! +...+ 2013/2014!)

= 1/2! - [(3-1)/3! + (4-1)/4!+(4-1)/5! + ... + (2014-1)/2014!]

=1/2! - [(3/3! + 4/4! + ...+ 2014/2014!) - (1/3! + 1/4! +... + 1/2013! + 1/2014!)] 

Ta có: Với n là số nguyên dương, n>2
\(\dfrac{n}{n!}\)=\(\dfrac{n}{1....\left(n-1\right)\left(n\right)}=\dfrac{1}{1.2....\left(n-1\right)}=\dfrac{1}{\left(n-1\right)!}\)

Do đó

A=1/2! - [ (1/2! + 1/3! + ... + 1/2013!) - (1/3!+ 1/4! +... + 1/2013! + 1/2014!) ]

= 1/2! - (1/2! - 1/2014!)

= 1/2014!

Vậy đáp án là A = \(\dfrac{1}{2014!}\)

Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)

\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)

\(=BC+C-BC-B\)

=C-B

\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)

\(A=\dfrac{2013}{1}+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}\)

\(=\left(\dfrac{2012}{2}+1\right)+\left(\dfrac{2011}{3}+1\right)+...+\left(\dfrac{1}{2013}+1\right)+1\)

\(=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)

\(=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)

\(P=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2014}=2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)

\(\Rightarrow\dfrac{P}{A}=\dfrac{2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)}{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)}=\dfrac{2013}{2014}\)

Vậy \(\dfrac{P}{A}=\dfrac{2013}{2014}\)

Lời giải:

Áp dụng BĐT AM-GM:

\(a^{2014}+\underbrace{1+1+....+1}_{1006}\geq 1007\sqrt[1007]{a^{2014}}=1007a^2\)

\(\Leftrightarrow a^{2014}+1006\geq 1007a^2\)

\(\Rightarrow a^{2014}+2013\geq 1007(a^2+1)\)

\(\Rightarrow \frac{a^{2014}+2013}{b^2+1}\geq \frac{1007(a^2+1)}{b^2+1}\). Hoàn toàn TT với các phân thức còn lại và cộng theo vế:

\(A\geq 1007\left(\frac{a^2+1}{b^2+1}+\frac{b^2+1}{c^2+1}+\frac{c^2+1}{a^2+1}\right)\)

\(\geq 1007.3\sqrt[3]{\frac{(a^2+1)(b^2+1)(c^2+1)}{(b^2+1)(c^2+1)(a^2+1)}}=3021\) (theo AM-GM)

Vậy \(A_{\min}=3021\Leftrightarrow a=b=c=1\)

\(\dfrac{2013}{2013+2014}< \dfrac{2013}{2013+2013}=\dfrac{1}{2}\)

Tương tự cộng theo vế suy ra đpcm

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