a,b,d áp dụng công thức này :
\(ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}\)
c)
\(x^2+y^2-4\left(x+y\right)=16=x^2-4x+y^2-4y+16\\ =\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\\ =\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-2=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\Leftrightarrow x=y=2\)
vậy \(MIN_C=8\) tại x = y = 8
a ) \(A=x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy Min A là : \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
b ) \(B=x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy Min B là : \(\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
c ) \(C=x^2+y^2-4\left(x+y\right)+16\)
\(=\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\)
\(=\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy Min C là : \(8\Leftrightarrow x=y=2\)
d ) \(D=2x^2+8x+9\)
\(=2\left(x^2+4x+4\right)+1\)
\(=2\left(x+2\right)^2+1\ge1\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy Min D là : \(1\Leftrightarrow x=-2\)