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 \(x=\dfrac{1}{2}\)

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

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

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

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

a: Khi x=2/3 thì \(A=\dfrac{\dfrac{2}{3}-2}{\dfrac{2}{3}}=\dfrac{-4}{3}\cdot\dfrac{3}{2}=-2\)

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

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

phần c đâu bạn

x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1 
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x) 
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1] 
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0 
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4 
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4 
Vay gia tri nho nhat P=4 khi x=1 
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4] 
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2 
Vay gia tri nho nhat Q= -9/2 khi x= 3/2 
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4 
= ( x-1/2)^2 + (y+3)^2 +3/4 
M>= 3/4 
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3 
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7] 
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7 
Vay GTLN A=7 khi x=2 
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4] 
GTLN B= 1/4 khi x=1/2 
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4) 
= -2[(x-1/2)^2 +9/4] 
GTLN N= -9/2 khi x=1/2

a) A = x² + 4x - 2 = x² + 4x + 4 - 6 = (x + 2)² - 6

(x + 2)² ≥ 0 => A ≥ -6 => GTNN của A là -6, xảy ra khi x = 2

`a)A=x^2+4x-2`

   `A=x^2+4x+4-6=(x+2)^2-6`

Vì `(x+2)^2 >= 0 AA x`

`<=>(x+2)^2-6 >= -6 AA x`

   Hay `A >= -6 AA x`

Dấu "`=`" xảy ra`<=>(x+2)^2=0<=>x=-2`

Vậy `GTN N` của `A` là `-6` khi `x=-2`

________________________________________________

`b)B=2x^2-4x+3`

   `B=2(x^2-2x+3/2)`

   `B=2(x^2-2x+1)+1=2(x-1)^2+1`

Vì `2(x-1)^2 >= 0 AA x`

`<=>2(x-1)^2+1 >= 1 AA x`

  Hay `B >= 1 AA x`

Dấu "`=`" xảy ra `<=>(x-1)^2=0<=>x=1`

Vậy `GTN N` của `B` là `1` khi `x=1`

__________________________________________________

`c)C=x^2+y^2-4x+2y+5`

   `C=x^2-4x+4+y^2+2y+1`

   `C=(x-2)^2+(y+1)^2`

Vì `(x-2)^2 >= 0 AA x` và `(y+1)^2 >= 0 AA y`

  `=>(x-2)^2+(y+1)^2 >= 0 AA x,y`

 Hay `C >= 0 AA x,y`

Dấu "`=`" xảy ra`<=>{((x-2)^2=0),((y+1)^2=0):}`

                         `<=>{(x=2),(y=-1):}`

Vậy `GTN N` của `C` là `0` khi `x=2`,y=-1