a,\(A=x^2+5x=x^2+2.\dfrac{5}{2}x\)
\(=x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}-\dfrac{25}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{25}{4}\)
Do\(\left(x+\dfrac{5}{2}\right)^2\ge0\) \(\left(\forall x\right)\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\)
Vậy Min A = \(\dfrac{-25}{4}\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)b, Sửa lại đề nha bn:
\(x^2-4xy+4y^2+10x-20y+28\)
\(=\left(x^2-4xy+4y^2+10x-20y+25\right)+3\)\(=\left(x-2y+5\right)^2+3\)
Ta có: \(\left(x-2y+5\right)^2\ge0\left(\forall x;y\right)\)
\(\Rightarrow\left(x-2y+5\right)^2+3\ge3\left(\forall x;y\right)\)
Vậy Min B = 3
Đúng 0
Bình luận (0)
