\(A=2x-2x^2-5\)

\(A=-2\left(x^2-x\right)-5\)

\(A=-2\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}-5\)

\(A=-2\left(x-\frac{1}{2}\right)^2-4\frac{1}{2}\)

Có \(2\left(x-\frac{1}{2}\right)^2\ge0\)với mọi x

=> \(-2\left(x-\frac{1}{2}\right)^2\le0\)với mọi x

=> \(-2\left(x-\frac{1}{2}\right)^2-4\frac{1}{2}\le-4\frac{1}{2}\)với mọi x

=> \(A\le-4\frac{1}{2}\)với mọi x

Dấu "=" xảy ra <=> \(x-\frac{1}{2}=0\)<=> \(x=\frac{1}{2}\)

KL: \(A_{max}=-4\frac{1}{2}\)<=> \(x=\frac{1}{2}\)

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) 

`f(x)=3x^2+x+x^4-x^3-x^2+2x+3`

`=x^4-x^3+2x^2+3x+3`

`g(x)=x^4+2x^2+x^3=x^4+x^3+2x^2`

b)

Bậc của `f(x)`: 4

Bậc của `g(x)`: 4

c)

`h(x)=f(x)+g(x)=x^4-x^3+2x^2+3x+3+x^4+x^3+2x^2`

`=2x^4+4x^2+3x+3`

`k(x)=g(x)-f(x)=x^4+x^3+2x^2-(x^4-x^3+2x^2+3x+3)`

`=x^4+x^3+2x^2-x^4+x^3-2x^2-3x-3`

`=2x^3-3x-3`

a

\(N=x-x^2\)

\(\Leftrightarrow-N=x^2-x\)

\(\Leftrightarrow-N+\frac{1}{4}=x^2-x+\frac{1}{4}\)

\(\Leftrightarrow-N+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

\(\Leftrightarrow-N=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\Rightarrow N_{max}=-\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

\(N=x-x^2\)

\(=-x^2+2.x.\frac{1}{2}-\frac{1}{4}+\frac{1}{4}\)

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

Vì \(-\left(x-\frac{1}{2}\right)^2\le0;\forall x\)

\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le0+\frac{1}{4};\forall x\)

Hay \(N\le\frac{1}{4};\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\)

                        \(\Leftrightarrow x=\frac{1}{2}\)

Vậy MAX \(N=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

\(N=x-x^2\)

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

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

\(=\left(-\frac{1}{4}+x\right)\left(\frac{3}{4}-x\right)\)

giải câu b trc nha

= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009

vậy min=2009 khi x=1

https://olm.vn//hoi-dap/question/57101.html     

Tham khảo đây nhá bạn

Ta có : P = x² - 2x + 5 = x² - 2x + 1 + 4 = (x - 1)² + 4

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

Suy ra : \(P=\left(x-1\right)^2+4\ge4\forall x\)

Nên : P_min = 4 khi x = 1

b) Ta có Q = 2x² - 6x = 2(x² - 3x) = 2(x² - 3x + \(\frac{9}{4}-\frac{9}{4}\) ) = \(2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)

Vì \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\) 

SUy ra ; \(Q=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Vậy \(Q_{min}=-\frac{9}{2}\) khi \(x=\frac{3}{2}\)

2−2x+5

2+2x−5)

2+2x+1)+6

2+6

2≤0∀x

2+6≤6∀x

Dấu "=" xảy ra ⇔

$x=-1$

Vậy $MA{X}_M=6\iff x=-1$

2+3

22−4x−3)

2−4x+4−7)

2−7]

2+7

22+7≤7

2=0⇔x=2

Vậy MAXA=7 khi x = 2

mới chỉ có GTLN thôi XD

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

b.\(B=x-x^2=\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)

c.\(C=2x-2x^2-5=-\frac{9}{2}-2\left(x^2-x+\frac{1}{4}\right)=-\frac{9}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{9}{2}\)