Đề hình như vậy mới đúng nè:
\(\left(x+5\right)^4+\left(x-4\right)^4=\left(2x+1\right)^4\)
~~~~ Bài làm ~~~~
Đặt: \(\left\{{}\begin{matrix}x+5=y\\x-4=z\end{matrix}\right.\Rightarrow2x+1=y+z\)
Ta có pt mới: \(y^4+z^4=\left(y+z\right)^4\)
\(\Leftrightarrow y^4+z^4=y^4+4y^3z+6y^2z^2+4yz^3+z^4\)
\(\Leftrightarrow2yz\left(2y^2+3yz+2z^2\right)=0\)
\(\Leftrightarrow2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\left(1\right)\)
Dễ thấy: \(7x^2+7x+22=7\left(x+\frac{1}{2}\right)^2+\frac{81}{4}>0\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-4=0\end{matrix}\right.\Leftrightarrow S=\left\{4;-5\right\}\)
Vậy ......
