Đặt \(x+2=a\)
\(\Rightarrow\left(a-1\right)^4+\left(a+1\right)^4=16\)
\(\Leftrightarrow a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=16\)
\(\Leftrightarrow2a^4+12a^2+2=16\)
\(\Leftrightarrow a^4+6a^2-7=0\Rightarrow\left[{}\begin{matrix}a^2=1\Rightarrow a=\pm1\\a^2=-7\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Đặt: x+1=t
=>t^4+(t+2)^4=16
\(\Leftrightarrow t^4+\left(t^2+4t+4\right)\left(t^2+4t+4\right)=16\Leftrightarrow t^4+t^4+4t^3+4t^2+4t^3+16t^2+16t+4t^2+16t+16=16\Leftrightarrow2t^4+8t^3+24t^2+32t+16=16\Leftrightarrow2t^4+8t^3+24t^2+32t=0\Leftrightarrow t^4+4t^3+24t^2+32t=0\Leftrightarrow\)
đi hok đã :3
