\(\left(x-2\right)^4+\left(x-3\right)^4=1\)\(\Leftrightarrow x^4-4x^3.2+6x^2.4-4x.8+16+x^4-4x^3.3+6x^2.9-4x.27+81=1\)
\(\Leftrightarrow2x^4-20x^3+78x^2-140x+96=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(2x^2-10x+16\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\\2x^2-10x+16=0\left(loại\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={2;3}
\(\left(x-2,5+0,5\right)^4+\left(x-2,5-0,5\right)^4=1\)
\(\left(t+\dfrac{1}{2}\right)^4+\left(t-\dfrac{1}{2}\right)^4=1\) với ( x -2,5 = t )
\(t^4+2t^3+\dfrac{3t^2}{2}+\dfrac{t}{2}+\dfrac{1}{16}+t^4-2t^3+\dfrac{3t^2}{2}-\dfrac{t}{2}+\dfrac{1}{16}=1\)
\(2t^4+3t^2-\dfrac{7}{8}=0\Leftrightarrow t^2=\dfrac{1}{4}\)
\(t=-\dfrac{1}{2};t=\dfrac{1}{2}.\)
+ x - 5/2 = -1/2 => x = 2
+ x -5/2 = 1/2 => x = 3
Vậy x = 2 hoặc x =3 .
