Question

giaỉ phương trình như sau

(x+5)⁴ + (x-4)⁴  = (2x+1)⁴

 

Answer 1

Đặt \(\hept{\begin{cases}x+5=y\\x-4=z\end{cases}}\)

\(\Leftrightarrow2x+1=y+z\)

=> PT có dạng

\(y^4+z^4=\left(y+z\right)^4\)

\(\Rightarrow y^4+z^4=y^4+4y^3z+6y^2z^2+4yz^3+z^4\)

\(\Leftrightarrow2yz\left(2y^2+3yz+2z^2\right)=0\)

\(\Leftrightarrow2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)

Dễ thấy  \(7x^2+7x+22=7\left(x+\frac{1}{2}\right)^2+\frac{81}{4}>0\)

Từ \(2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=4\end{cases}}\)

Vậy .....