a,Ta có :\(A=x\left(x-6\right)=x^2-6x\)
\(=x^2-6x+9-9\)
\(=\left(x-3\right)^2-9\)
Vì: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\)\(\left(x-3\right)^2-9\ge-9\forall x\)
Hay: \(A\ge-9\forall x\)
Dấu = xảy ra khi (x-3)^2=0
<=>x=3
Vậy Min A= -9 tại x=3
b,Ta có: \(B=-3x\left(x+3\right)-7\)
\(=-3x^2-9x-7\)
\(=-3\left(x^2+3x+\frac{7}{3}\right)\)
\(=-3\left[\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{12}\right]\)
\(=-3\left[\left(x+\frac{3}{2}\right)^2+\frac{1}{12}\right]\)
\(=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\)
Vì: \(-3\left(x+\frac{3}{2}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le\frac{-1}{4}\forall x\)
Hay \(B\le\frac{-1}{4}\forall x\)
Dấu = xảy ra khi \(-3\left(x+\frac{3}{2}\right)^2=0\)
\(\Rightarrow x=\frac{-3}{2}\)
Vậy Max B=-1/4 tại x=-3/2
a) \(A=x\left(x-6\right)=x^2-6x+9-9=\left(x-3\right)^2-9\ge-9\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=3\)
Vậy Min A = -9 khi x = 3
b) \(B=-3x\left(x+3\right)-7=-3x^2-9x-7=-3\left(x^2+9x+20,25\right)+53,75\)
\(=-3\left(x+4,5\right)^2+53,75\le53,75\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-4,5\)
Vậy Max B = 53,75 khi x = -4,5
câu b mk lm nhầm, bn tham khảo của MIYANO SHINO nhé