Câu 2:
f(3)=f(-3)
=>\(a\cdot3^2+b\cdot3+c=a\cdot\left(-3\right)^2+b\cdot\left(-3\right)+c\)
=>3b=-3b
=>6b=0
=>b=0
Vậy: \(f\left(x\right)=ax^2+c\)
\(f\left(-x\right)=a\cdot\left(-x\right)^2+c=a\cdot x^2+c\)
=>f(x)=f(-x)
Câu 3:
\(\left|x-2\right|>=0\forall x\)
\(\left(2y-1\right)^2>=0\forall x\)
\(z^2>=0\forall z\)
=>\(z^2+1>=1\forall z\)
=>\(\left(z^2+1\right)^4>=1\forall z\)
=>\(\left|x-2\right|+\left(2y-1\right)^2+\left(z^2+1\right)^4>=1\forall x,y,z\)
=>\(\left|x-2\right|+\left(2y-1\right)^2+\left(z^2+1\right)^4+5>=1+5=6\forall x,y,z\)
=>\(A>=6\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\2y-1=0\\z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\\z=0\end{matrix}\right.\)
