\(\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge\frac{\left(a+b\right)^2}{a+b-2}=8+\frac{\left(a+b-4\right)^2}{a+b-2}\ge8\)

\(n^4+2n^3+2n^2+n+7=k^2\)

\(\Leftrightarrow\left(n^2+n\right)^2+\left(n^2+n\right)+7=k^2\)

\(\Leftrightarrow4\left(n^2+n\right)^2+4\left(n^2+n\right)+1+27=4k^2\)

\(\Leftrightarrow\left(2n^2+2n+1\right)^2-4k^2=-27\)

\(\Leftrightarrow\left(2n^2+2n+1-2k\right)\left(2n^2+2n+1+2k\right)=-27\)

Làm nôt

Dễ thây \(\hept{\begin{cases}\sqrt{a^2+2017}-a\ne0\\\sqrt{b^2+2017}-b\ne0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(a+\sqrt{a^2+2017}\right)\left(\sqrt{a^2+2017}-a\right)\left(b+\sqrt{b^2+2017}\right)=2017\left(\sqrt{a^2+2017}-a\right)\\\left(a+\sqrt{a^2+2017}\right)\left(b+\sqrt{b^2+2017}\right)\left(\sqrt{b^2+2017}-b\right)=2017\left(\sqrt{b^2+2017}-b\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2017\left(b+\sqrt{b^2+2017}\right)=2017\left(\sqrt{a^2+2017}-a\right)\\2017\left(a+\sqrt{a^2+2017}\right)=2017\left(\sqrt{b^2+2017}-b\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b+\sqrt{b^2+2017}=\sqrt{a^2+2017}-a\\a+\sqrt{a^2+2017}=\sqrt{b^2+2017}-b\end{cases}}\)

\(\Leftrightarrow a+b=0\)

Ta có : \(\sqrt{x^4}=\sqrt{x^2.x^2}=\sqrt{x^2}.\sqrt{x^2}=x.x=x^2=7\)

\(\Rightarrow x=\sqrt{7}\)

Vậy \(x=\sqrt{7}\)

\(\sqrt{x^4}=7\)

\(\Rightarrow x^2=7\)

\(\Rightarrow x^2-7=0\)

\(\Rightarrow\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-\sqrt{7}=0\\x+\sqrt{7}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\)

Vậy....

\(x^2-3xy+y^2+y^3=0\)

\(\Rightarrow\Delta_x=9y^2-4\left(y^3+y^2\right)\ge0\)

\(\Rightarrow0\le y< 2\)

\(\Rightarrow y=\left\{0;1\right\}\)

Làm nôt

\(z=\frac{P-x-y}{2}\)

\(\Rightarrow x^2+y^2+\frac{\left(P-x-y\right)^2}{4}=3\)

\(\Leftrightarrow5y^2+\left(2x-2P\right)y+5x^2-2Px+P^2-12=0\)

\(\Rightarrow\Delta_y=\left(x-P\right)^2-5.\left(5x^2-2Px+P^2-12\right)\ge0\)

\(\Leftrightarrow36x^2-12Px+P^2+5P^2-90\le0\)

\(\Leftrightarrow5P^2-90\le-\left(6x-P\right)^2\le0\)

\(\Leftrightarrow-3\sqrt{2}\le P\le3\sqrt{2}\)

\(\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)

Tương tụ co:

\(\hept{\begin{cases}\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\\\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\end{cases}}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Leftrightarrow xyz\le\frac{1}{8}\)