a,\(A=-\left(x^3-5x+7\right)=-\left(x^2-5x+\dfrac{25}{4}+\dfrac{3}{4}\right)=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\)vậy A luôn âm với mọi x
b,\(B=x^2+6x+\left|y-5\right|-6\)
với \(y\ge5\)
\(=>B=x^2+6x+y-5-6=x^2+6x+y-11\)
\(=B\ge x^2+6x+5-11=x^2+6x+9-15=\left(x+3\right)^2-15\ge-15\)
dấu"=" xảy ra<=>x=-3,y=5
a) Ta có: \(A=-x^2+5x-7\)
\(=-\left(x^2-5x+7\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)(đpcm)
b) Ta có: \(B=x^2+6x+\left|y-5\right|-6\)
\(=x^2+6x+9+\left|y-5\right|-15\)
\(=\left(x+3\right)^2+\left|y-5\right|-15\ge-15\forall x,y\)
Dấu '=' xảy ra khi x=-3 và y=5
