Vì \(\left(x+1\right)^4\ge0\forall x\); \(\left(x-3\right)^4\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^4+\left(x-3\right)^4\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}\left(ktm\right)}\)
=> Pt vô nghiệm
a) ( x + 1 ) 4 + ( x - 3 ) 4 = 0
Vì \(\left(x+1\right)^4\ge0\forall x\inℤ\)
\(\left(x-3\right)^4\ge0\forall x\inℤ\)
Nên \(\left(x+1\right)^4+\left(x-3\right)^4=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=3\end{cases}}}\)
Vậy .....
( x + 1 )4 + ( x - 3 )4 = 0
\(\hept{\begin{cases}\left(x+1\right)^4\\\left(x-3\right)^4\end{cases}}\ge0\forall x\Rightarrow\left(x+1\right)^4+\left(x-3\right)^4\ge0\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\)( mâu thuẫn )
=> Pt vô nghiệm
