Có ab > 2013a + 2014b <=> 1 > 2013/b + 2014/a (vì a,b >0 )

\(\Leftrightarrow a+b>\frac{2013\left(a+b\right)}{b}+\frac{2014\left(a+b\right)}{a}=2013+2014+\frac{2013a}{b}+\frac{2014b}{a}\)

Mà \(\frac{2013a}{b}+\frac{2014b}{a}\ge2\sqrt{2013\cdot2014}\)

\(\Rightarrow a+b>2013+2014+2\sqrt{2013\cdot2014}=\left(\sqrt{2013}+\sqrt{2014}\right)^2\)

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Tích cho mk nhoa !!!! ~~~

\(4\sqrt[4]{a}+7\sqrt[7]{b}\ge11\sqrt[11]{ab}\)

Bài 3. 

\(\left\{{}\begin{matrix}a\left(a+b+c\right)=-\dfrac{1}{24}\left(1\right)\\c\left(a+b+c\right)=-\dfrac{1}{72}\left(2\right)\\b\left(a+b+c\right)=\dfrac{1}{16}\left(3\right)\end{matrix}\right.\)

Dễ thấy \(a,b,c\ne0\Rightarrow a+b+c\ne0\)

Chia (1) cho (2), ta được \(\dfrac{a}{c}=3\Rightarrow a=3c\left(4\right)\)

Chia (2) cho (3) ta được: \(\dfrac{c}{b}=-\dfrac{2}{9}\Rightarrow b=-\dfrac{9}{2}c\left(5\right)\).

Thay (4), (5) vào (2), ta được: \(-\dfrac{1}{2}c^2=-\dfrac{1}{72}\)

\(\Rightarrow c=\pm\dfrac{1}{6}\).

Với \(c=\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=\dfrac{1}{2}\\b=-\dfrac{9}{2}c=-\dfrac{3}{4}\end{matrix}\right.\)

Với \(c=-\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=-\dfrac{1}{2}\\b=-\dfrac{9}{2}c=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(\left(a;b;c\right)=\left\{\left(\dfrac{1}{2};-\dfrac{3}{4};\dfrac{1}{6}\right);\left(-\dfrac{1}{2};\dfrac{3}{4};-\dfrac{1}{6}\right)\right\}\)

Ta có : \(\frac{a}{a+\sqrt{2013a+bc}}=\frac{a}{a+\sqrt{a^2+ab+ac+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Theo bất đẳng thức Bunhiacopxki : \(\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ab}+\sqrt{ac}\)

\(\Rightarrow\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

hay \(\frac{a}{a+\sqrt{2013a+bc}}\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Tương tự : \(\frac{b}{b+\sqrt{2013b+ac}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(\frac{c}{c+\sqrt{2013c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng các bất đẳng thức trên theo vế được \(\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ac}}+\frac{c}{c+\sqrt{2013c+ab}}\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\\a+b+c=2013\\a,b,c>0\end{cases}}\) \(\Leftrightarrow a=b=c=671\)

v

Ta có: \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\)

\(\Leftrightarrow\left(x-\sqrt{x^2+2013}\right)\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-y-\sqrt{y^2+2013}=x-\sqrt{x^2+2013}\)

⇔\(x+y=\sqrt{x^2+2013}-\sqrt{y^2+2013}\)(1)

Nhân liên hợp tương tự nhân \(y-\sqrt{y^2+2013}\)vào hai về rút được

\(x+y=\sqrt{y^2+2013}-\sqrt{x^2+2013}\)(2)

Cộng vế theo vế (1)(2) ta được \(x+y=0\Rightarrow x=-y\)

Thay vào \(A=\left(-y\right)^{2014}-y^{2014}+1=1\)

sửa lại đề đi

c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)

\(=4+\sqrt{10}-4+\sqrt{10}\)

\(=2\sqrt{10}\)

d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)

\(=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1\)

\(=2\sqrt{2}\)

a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)

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

c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)

d) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)

b, Ta có \(2015^2=\left(2014+1\right)^2=2014^2+2.2014+1\) 

=> \(2014^2+1=2015^2-2.2014\) 

=> \(B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

= \(\sqrt{2015^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\) 

= \(\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\) = \(2015-\frac{2014}{2015}+\frac{2014}{2015}=2015\) 

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