Đặt x+3=a
Có :\(\left(a-2\right)^2+\left(a-1\right)^3+a^4=2\)
<=> \(a^4+a^3-3a^2+3a-1+a^2-4a+4-2=0\)
<=> \(a^4+a^3-2a^2-a+1=0\)
<=> \(\left(a^4-2a^2+1\right)+a\left(a^2-1\right)=0\)
<=> \(\left(a^2-1\right)^2+a\left(a^2-1\right)=0\)
<=>\(\left(a^2-1\right)\left(a^2+a-1\right)=0\)
=> \(\left[{}\begin{matrix}a^2=1\\a^2+a-1=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}a=1\\a=-1\\\left(a+\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x+3=1\\x+3=-1\\a+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-2\\x=-4\\a=\frac{\sqrt{5}-1}{2}\\a=-\frac{1+\sqrt{5}}{2}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-2\\x=-4\\x+3=\frac{\sqrt{5}-1}{2}\\x+3=-\frac{1+\sqrt{5}}{2}\end{matrix}\right.\)
<=> \(\)\(\left[{}\begin{matrix}x=-2\\x=-4\\x=\frac{\sqrt{5}-7}{2}\\x=-\frac{\sqrt{5}+7}{2}\end{matrix}\right.\)(t/m)
\(\left[{}\begin{matrix}x=-2\\x=-4\\x=\frac{\sqrt{5}-7}{2}\\x=-\frac{\sqrt{5}+7}{2}\end{matrix}\right.\)\(\left[{}\begin{matrix}x=-2\\x=-4\\x=\frac{\sqrt{5}-7}{2}\\x=-\frac{\sqrt{5}+7}{2}\end{matrix}\right.\)
