Dat \(\left\{{}\begin{matrix}x-2=a\\3-x=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\a^4+b^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\left(a^2+b^2\right)^2-2a^2b^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\a^2b^2-2ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1-b\left(1\right)\\ab\left(ab-2\right)=0\left(2\right)\end{matrix}\right.\)
(2)
PT(2)
\(\Leftrightarrow\left[{}\begin{matrix}ab=0\\ab=2\end{matrix}\right.\)
Thay (1) vao (2)
\(\Leftrightarrow\left[{}\begin{matrix}\left(1-b\right)b=0\left(3\right)\\\left(1-b\right)b=2\left(4\right)\end{matrix}\right.\)
(3)\(\Leftrightarrow\left[{}\begin{matrix}b=0\\b=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3-x=0\\3-x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Tuong tu may cai kia la xong
Cách khác:
Đặt $x-2,5=a$. Khi đó $x-2=a+0,5$ và $x-3=a-0,5$
PT trở thành:
\((a+0,5)^4+(a-0,5)^4=1\)
\(\Leftrightarrow 2a^4+3a^2+0,125=1\)
\(\Leftrightarrow 2a^4+3a^2-0,875=0\)
Coi $a^2=t (t\geq 0)$ thì PT trở thành: $2t^2+3t-0,875=0$
Đây là PT bậc 2 một ẩn, ta dễ dàng giải ra $t=\frac{1}{4}$
\(\Rightarrow a^2=\frac{1}{4}\Rightarrow a=\pm \frac{1}{2}\)
\(\Rightarrow x=2\) hoặc $x=3$
