\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

um

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

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

Giups mik vs

a)\(A=2x+1-x^2=2-\left(x^2-2x+1\right)=2-\left(x-1\right)^2\le2;\forall x\)

\(\Rightarrow A_{max}=2\Leftrightarrow x=1\)

b)\(B=4x-4x^2-5=-4-\left(4x^2-4x+1\right)=-4-\left(2x-1\right)^2\le-4;\forall x\)

\(\Rightarrow B_{max}=-4\Leftrightarrow x=\dfrac{1}{2}\)

a) `A=2x+1-x^2`

`=-(x^2-2x-1)`

`=-(x^2-2x+1)+2`

`=-(x-1)^2+2`

Có: `-(x-1)^2 <= forall x => -(x-1)^2+2 <=2`

`=> A_(max)=2 <=> x=1`

b) `B=4x-4x^2-5`

`=-(4x^2-4x+5)`

`=-(4x^2-4x+1)-4`

`=-[(2x)^2-2.2x.1+1^2]-4`

`=-(2x-1)^2+4`

`=> B_(max)=4 <=> x=1/2`

a) Ta có: \(A=-x^2+2x+1\)

\(=-\left(x^2+2x-1\right)\)

\(=-\left(x+1\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-1

b) Ta có: \(B=-4x^2+4x-5\)

\(=-\left(4x^2-4x+5\right)\)

\(=-\left(2x-1\right)^2-4\le-4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a) Giá trị lớn nhất:

\(A=2x-3x^2-4=-3\left(x^2-\frac{2}{3}x+\frac{4}{3}\right)=-3\left[x^2-2.x.\frac{1}{3}+\left(\frac{1}{3}\right)^2+\frac{35}{9}\right]=-3\left(x-\frac{1}{3}^2\right)-\frac{35}{3}\)

Vì \(\left(x-\frac{1}{3}\right)^2\ge0\left(x\in R\right)\)

Nên \(-3\left(x-\frac{1}{3}\right)^2\le0\left(x\in R\right)\)

do đó \(-3\left(x-\frac{1}{3}\right)^2-\frac{35}{3}\le-\frac{35}{3}\left(x\in R\right)\)

Vậy \(Max_A=-\frac{35}{3}\)khi \(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)

\(B=-x^2-4x=-\left(x^2+4x\right)=-\left(x^2+2.x.2+2^2-2^2\right)=-\left(x+2\right)^2+4\)

Vì \(\left(x+2\right)^2\ge0\left(x\in R\right)\)

nên \(-\left(x+2\right)^2\le0\left(x\in R\right)\)

do đó \(-\left(x+2\right)^2+4\le4\left(x\in R\right)\)

Vậy \(Max_B=4\)khi \(x+2=0\Rightarrow x=-2\)

b) Giá trị nhỏ nhất 

\(A=x^2-2x-1=x^2-2.x.+1-2=\left(x-1\right)^2-2\)

Vì \(\left(x-1\right)^2\ge0\left(x\in R\right)\)

nên \(\left(x-1\right)^2-2\ge-2\left(x\in R\right)\)

Vậy \(Min_A=-2\)khi \(x-1=0\Rightarrow x=1\)

\(B=4^2+4x+5=\left(2x\right)^2+2.2x.1+1+4=\left(2x+1\right)^2+4\)

vì \(\left(2x+1\right)^2\ge0\left(x\in R\right)\)

nên \(\left(2x+1\right)^2+4\ge4\left(x\in R\right)\)

Vậy \(Min_B=4\)khi \(2x+1=0\Rightarrow x=-\frac{1}{2}\)

a: Ta có: \(A=-x^2+4x+3\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7\le7\forall x\)

Dấu '=' xảy ra khi x=2

b: Ta có: \(B=-x^2+x\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Câu d đề sai bạn nhé, biểu thức này chỉ có min, không có max