Ta có: \(\left(4,5-x\right)^4+\left(5,5-x\right)^4=1\)
=>\(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)
=>\(\left\lbrack\left(x-5\right)+0,5\right\rbrack^4+\left\lbrack\left(x-5\right)-0,5\right\rbrack^4=1\) (1)
Đặt a=x-5; b=0,5
\(\left(a+b\right)^4+\left(a-b\right)^4\)
\(=\left(a^2+2ab+b^2\right)^2+\left(a^2-2ab+b^2\right)^2\)
\(=\left\lbrack\left(a^2+b^2\right)^2+2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack+\left\lbrack\left(a^2+b^2\right)^2-2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack\)
\(=2\left(a^4+2a^2b^2+b^4\right)+8a^2b^2=2a^4+12a^2b^2+2b^4\)
\(=2\left(x-5\right)^4+12\cdot\left(x-5\right)^2\cdot\left(0,5\right)^2+2\cdot\left(0,5\right)^4\)
\(=2\left(x-5\right)^4+3\left(x-5\right)^2+0,125\)
(1) sẽ trở thành: \(2\left(x-5\right)^4+3\left(x-5\right)^2+0,125=1\)
=>\(2\left(x-5\right)^4+3\left(x-5\right)^2-0,875=0\)
=>\(16\left(x-5\right)^4+24\left(x-5\right)^2-7=0\)
=>\(16\left(x-5\right)^4+28\left(x-5\right)^2-4\left(x-5\right)^2-7=0\)
=>\(\left\lbrack4\left(x-5\right)^2+7\right\rbrack\left\lbrack4\left(x-5\right)^2-1\right\rbrack=0\)
=>\(4\left(x-5\right)^2-1=0\)
=>\(\left(2x-10\right)^2=1\)
=>\(\left[\begin{array}{l}2x-10=1\\ 2x-10=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=11\\ 2x=9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}\\ x=\frac92\end{array}\right.\)
