a/ Với \(x=\left\{2;3\right\}\) là nghiệm
- Với \(x>3\Rightarrow\left\{{}\begin{matrix}x-3>0\\x-2>1\end{matrix}\right.\) \(\Rightarrow VT>1\Rightarrow VT>VP\)
Phương trình vô nghiệm
- Với \(x< 2\) viết lại pt: \(\left(2-x\right)^4+\left(3-x\right)^4=1\)
\(\left\{{}\begin{matrix}2-x>0\\3-x>1\end{matrix}\right.\) \(\Rightarrow VT>1\Rightarrow VT>VP\Rightarrow\) ptvn
- Với \(2< x< 3\) viết lại pt: \(\left(x-2\right)^4+\left(3-x\right)^4=1\)
\(\left\{{}\begin{matrix}0< x-2< 1\\0< 3-x< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^4< x-2\\\left(3-x\right)^4< 3-x\end{matrix}\right.\)
\(\Rightarrow VT< x-2+3-x\Rightarrow VT< 1\Rightarrow VT< VP\Rightarrow\) ptvn
Vậy pt có 2 nghiệm: \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b/ Đặt \(\left\{{}\begin{matrix}2x-5=a\\x+1=b\end{matrix}\right.\) \(\Rightarrow3x-4=a+b\)
\(a^3-\left(a+b\right)^3+b^3=0\)
\(\Leftrightarrow a^3+b^3-\left[a^3+b^3+3ab\left(a+b\right)\right]=0\)
\(\Leftrightarrow ab\left(a+b\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x+1=0\\3x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{matrix}\right.\)
