We put \(n^2-14n+38=k^2\)

\(\Rightarrow n^2-14n+49-11=k^2\)

\(\Rightarrow\left(n-7\right)^2-11=k^2\)

\(\Rightarrow\left(n-7\right)^2-k^2=11\)

\(\Rightarrow\left(n-7-k\right)\left(n-7+k\right)=11=1.11=11.1=\left(-1\right).\left(-11\right)\)

\(=\left(-11\right).\left(-1\right)\)

Prints:

Case by case, we have \(n\in\left\{13;1\right\}\)

Rút gọn hả bạn

\(=\frac{2\sqrt{2}\left(1-\sqrt{3}\right)}{3\sqrt{2-\sqrt{3}}}\)

\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2}.\sqrt{2+\sqrt{3}}}{3.\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}}\)

\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2\left(2+\sqrt{3}\right)}}{3.\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)

\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{4+2\sqrt{3}}}{3.\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)

\(=\frac{2\left(1-\sqrt{3}\right)\sqrt{\left(1+\sqrt{3}\right)^2}}{3.\sqrt{4-3}}\)

\(=\frac{2\left(1-\sqrt{3}\right)|1+\sqrt{3}|}{3\sqrt{1}}\)

\(=\frac{2\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}{3}\)

\(=\frac{2\left(1^2-\left(\sqrt{3}\right)^2\right)}{3}\)

\(=\frac{2.\left(-2\right)}{3}=\frac{-4}{3}\)

a) \(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng \(\forall a,b\) )

=>đpcm

Cô si

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2c\)

\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}=2a\)

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)

Cộng lại ta có:

\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\Rightarrowđpcm\)

Ta co:

 \(9=x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Rightarrow-3\sqrt{2}\le x+y\le3\sqrt{2}\)

Dat \(\hept{\begin{cases}a=x+y\\b=xy\end{cases}\left(a\ne-3,-3\sqrt{2}\le a\le3\sqrt{2}\right)}\)

\(\Rightarrow a^2-2b=9\Leftrightarrow\frac{a^2}{2}-\frac{9}{2}=b\) 

\(\Rightarrow Q=\frac{b}{a+3}=\frac{a^2-9}{2a+6}=\frac{a-3}{2}=\frac{x+y-3}{2}\)

Xet \(0\le x+y\le3\sqrt{2}\)

\(\Rightarrow Q=\frac{x+y-3}{2}\le\frac{\sqrt{2\left(x^2+y^2\right)}-3}{2}=\frac{3\sqrt{2}-3}{2}\)  

Dau '=' xay ra khi \(x=y=\frac{3}{\sqrt{2}}\)

Xet \(-3\sqrt{2}\le x+y< 0\)

\(\Rightarrow Q=\frac{x+y-3}{2}\ge\frac{-3\sqrt{2}-3}{2}\)

Dau '=' xay ra khi \(x=y=-\frac{3}{\sqrt{2}}\)