\(4x^4+4x^3+x^2+3x\ge0\)

\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)

\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)

• \(u+v+x^2-x+1=0\Leftrightarrow u+v+\left(x-\frac{1}{2}\right)^2=-\frac{3}{4}\)
• \(u=v\Leftrightarrow4x^4+4x^2+1=4x^4+4x^3+x^2+3x\Leftrightarrow\left(x-1\right)^3=-3x^3\Leftrightarrow x-1=-x\sqrt[3]{3}\Leftrightarrow x=\frac{1}{1+\sqrt[3]{3}}\)Đối chiếu điều kiện ta thu được nghiệm duy nhất \(x=\frac{1}{1+\sqrt[3]{3}}\)

\(5x^4+y^2-4x^2y-85=0\)

\(\left(2x^2\right)^2-2.2x^2.y+y^2+x^4=85\)

\(\left(2x^2-y\right)^2+x^4=85\)

Mà \(85=2^2+3^4=\left(-2\right)^2+\left(-3\right)^4\)

Vì phương trình nghiệm nguyên nên:

\(\left(2x^2-y\right)^2+x^4=2^2+3^4\)

\(\Rightarrow\orbr{\begin{cases}2x^2-y=2\\x=3\end{cases}}\)     hoặc      \(\orbr{\begin{cases}2x^2-y=3\\x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2.3^2-y=2\\x=3\end{cases}}\)   hoặc       \(\orbr{\begin{cases}2.2^2-y=3\\x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}18-y=2\\x=3\end{cases}}\)      hoặc         \(\orbr{\begin{cases}8-y=3\\x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=16\\x=3\end{cases}}\)                hoặc         \(\orbr{\begin{cases}y=5\\x=2\end{cases}}\)

Vậy..............