Ta có : \(\left(x+1\right)^4\ge0\forall x\)
\(\left(x+3\right)^4\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^4+\left(x+3\right)^4\ge0\forall x\)
Dấu = xảy ra khi : \(\left(x+1\right)^4+\left(x+3\right)^4=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x+3\right)^4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\x=-3\end{cases}\left(ktm\right)}\)
\(\Rightarrow\)phương trình vô ngiệm
Ta có :
\(\left(x+1\right)^4\ge0\forall x\)
\(\left(x+3\right)^4\ge0\forall x\)
Phương trình = 0 \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x+3\right)^4=0\end{cases}}\)
\(\hept{\begin{cases}x+1=0\\x+3=0\end{cases}}\)
\(\hept{\begin{cases}x=-1\\x=-3\end{cases}}\)
\(x\in\varnothing\)
Ta có:\(\orbr{\begin{cases}\left(x+1\right)^4\ge0\forall x\\\left(x+3\right)^4\ge0\forall x\end{cases}\Rightarrow\left(x+1\right)^4+\left(x+3\right)^4\ge0\forall x}\)
Dấu"="xảy ra khi\(\orbr{\begin{cases}\left(x+1\right)^4=0\\\left(x+3\right)^4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}}\)(mâu thuẫn)
=>Vô nghiệm
