\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra

thì A=\(\left|3-x\right|+\left|x-2\right|\ge\left|3-x+x-2\right|=1\) (bất đẳng thức về dâu giá trị tuyệt đối)

dấu = xảy ra <=> tích của chúng = nhau

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

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

a: =x^2-10x+25+y^2+2y+1

=(x-5)^2+(y+1)^2>=0

Dấu = xảy ra khi x=5 và y=-1

b: x^2-3x-2

=x^2-3x+9/4-17/4

=(x-3/2)^2-17/4>=-17/4

Dấu = xảy ra khi x=3/2

Ta có : \(P=3A+2B\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{3}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}.\)

\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+2\right)-1}{\sqrt{x}+2}=2-\dfrac{1}{\sqrt{x}+2}\)

Do \(x\ge0\Rightarrow\sqrt{x}+2\ge0\)

\(\Rightarrow-\dfrac{1}{\sqrt{x}+2}\ge-1\)

\(\Rightarrow P=2-\dfrac{1}{\sqrt{x}+2}\ge-1+2=1.\)

Vậy : \(MinP=1.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=0.\)