\(\left(x+4\right)^4+\left(x+6\right)^4=82\)

Đặt a = x + 5

Ta có:

\(\left(x+4\right)^4+\left(x+6\right)^4=82\)

\(\Leftrightarrow\left(a-1\right)^4+\left(a+1\right)^4\)

\(\Leftrightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=82\)

\(\Leftrightarrow\left(a^2-2a+1\right)^2+\left(a+2a+1\right)^2=82\)

\(\Leftrightarrow\left(a^2+1\right)^2-4a\left(a^2+1\right)+4a^2+\left(a^2+1\right)^2+4a\left(a^2+a\right)+4a^2=82\) \(\Leftrightarrow\left(a^2+1\right)^2+4a^2=41\)

\(\Leftrightarrow a^4+6a^2+1=41\)

\(\Leftrightarrow a^4+6a^2-40a=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a^2=-10\left(loại\right)\\a^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-7\end{matrix}\right.\)

Đặt \(x-1=a\) phương trình trở thành:

\(\left(a+2\right)^4+\left(a-2\right)^4=82\)

\(\Leftrightarrow a^4+8a^3+24a^2+32a+16+a^4-8a^3+24a^2-32a+16=82\)

\(\Leftrightarrow2a^4+48a^2+32=82\)

\(\Leftrightarrow a^4+24a^2-25=0\Rightarrow\left[{}\begin{matrix}a^2=1\\a^2=-25\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x-4\right)=\left(x^2-2x-3\right)\left(x^2-2x-8\right)\)

Đặt \(x^2-2x-3=t\)

\(\text{pt thành }t\left(t-5\right)=36\Leftrightarrow t^2-5t-36=0\Leftrightarrow t=9\text{ hoặc }t=-4\)

\(+t=9\Rightarrow x^2-2x-3=9\Leftrightarrow x^2-2x-12=0\Leftrightarrow x=1+\sqrt{13}\text{ hoặc }x=1-\sqrt{13}\)

\(+t=-4\Rightarrow x^2-2x-3=-4\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy ....

giúp