a: \(A=x^2-6x+10\)
\(=x^2-6x+9+1=\left(x-3\right)^2+1>=1\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
b: \(B=x^2+5y^2+2xy-2y+2025\)
\(=x^2+2xy+y^2+4y^2-2\cdot2y\cdot\dfrac{1}{2}+\dfrac{1}{4}+2024,75\)
\(=\left(x+y\right)^2+\left(2y-\dfrac{1}{2}\right)^2+2024,75>=2024,75\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\2y-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{4}\\x=-y=-\dfrac{1}{4}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
