`1)(2x^2+x+1)(2x^2+x+4)=-4`
`<=>4x^4+4x^3+11x^2+5x+4=-4`
`<=>4x^4+4x^3+11x^2+5x+4+4=-4+4`
`<=>4x^4+4x^3+11x^2+5x+4+4=0`
Vậy `x∈ZZ`
`2)4x^3-4x^2-15x+18=0`
`<=>(x+2)(2x-3)^2=0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\2x-3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-2\end{array} \right.\)
`3)x^4+2x^3-4x^2-5x-6=0`
`<=>x^4+x^3+x^2-2x^3-2x^2+2x+3x^3+3x^2+3x-6x^2+6x+6=0`
`<=>(x^2+x+1)(x^2-2x+3x-6)=0`
`<=>(x^2+x+1)(x-2)(x+3)`
Ta có : `x^2+x+1>0∀x` nên vô nghiệm
`<=>` \(\left[ \begin{array}{l}x+3=0\\x-2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right. \)`
`4)x^4-5x^3+7x^2-5x+1=0`
`<=>x^3-4.69847...x^2+5.58330...x-3.31651... ≈0`
`<=>x^2-1.38196...x+1.00000... ≈0`
`<=>x ≈0.30152... +1.00000≈0`
`5)(x+2)^4+(x+4)^4=16`
`<=>2x^4+24x^3+120x^2+288x+272=16`
`<=>2x^4+24x^3+120x^2+288x+272-16=16-16`
`<=>2x^4+24x^3+120x^2+288x+256=0`
`<=>2(x+2)(x+4)(x^2+6x+16)=0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\x+4=0\\x^2+6x+16=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=-4\\x=-3+\sqrt[]{7};-3-\sqrt[]{7}\end{array} \right.\)
