\(P=3-4x-x^2=-\left(x^2+4x+4\right)+7\)

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

\(Do-\left(x+2\right)^2\le0\Leftrightarrow P\le7\)

Dấu "=" xảy ra khi  x + 2 =0

    => x = -2 

Vậy Max P = 7 khi x = - 2

giúp mình câu 2 điii

b) \(Q=2x-2-3x^2\)

\(\Leftrightarrow Q=-\left[\left(3x^2-2x+\frac{1}{3}\right)+\frac{5}{3}\right]\)

\(\Leftrightarrow Q=-\left[\left(\sqrt{3}x-\frac{1}{\sqrt{3}}\right)^2+\frac{5}{3}\right]\le\frac{5}{3}\)

Dấu " = " xảy ra :

\(\Leftrightarrow\sqrt{3}x-\frac{1}{\sqrt{3}}=0\)

\(\Leftrightarrow\sqrt{3}x=\frac{1}{\sqrt{3}}\)

\(\Leftrightarrow x=\frac{1}{3}\)

Vậy \(Max_Q=\frac{5}{3}\Leftrightarrow x=\frac{1}{3}\)

c) \(R=2-x^2-y^2-2\left(x+y\right)\)

\(\Leftrightarrow R=-\left[\left(x^2+2x+1\right)+\left(y^2+2y+1\right)-4\right]\)

\(\Leftrightarrow R=-\left[\left(x+1\right)^2+\left(y+1\right)^2-4\right]\le-4\)

Dấu " = " xảy ra :

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\y+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)

Vậy \(Max_Q=-4\Leftrightarrow x=y=-1\)

d) \(S=-x^2+4x-9\)

\(\Leftrightarrow S=-\left[\left(x^2-4x+4\right)+5\right]\)

\(\Leftrightarrow S=-\left[\left(x-2\right)^2+5\right]\le5\)

Dấu " = " xảy ra :

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(Max_S=5\Leftrightarrow x=2\)

a)

\(Q=2x-2-3x^2\\ Q=-3\left(x-\dfrac{1}{3}\right)^2+\dfrac{3.\left(-3\right)\left(-2\right)-2^2}{4.\left(-3\right)}\\ Q=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{12}\le-\dfrac{14}{12}\)

đẳng thức xảy ra khi x=1/3

vậy MAX Q=-14/12 tại x=1/3

c)

\(S=-x^2+4x-9\\ S=-\left(x^2-4x+4\right)-5\\ S=-\left(x-2\right)^2-5\le-5\)

đẳng thức xảy ra khi x=2

vậy MAX S=-5 tại x=2

a) \(Q=2x-2-3x^2\)

\(=-\left(3x^2-2x+2\right)\)

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

\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{5}{9}\right)\)

\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\)

Ta có: \(-3\left(x-\dfrac{1}{3}\right)^2\le0\forall x\Rightarrow-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{5}{3}\le-\dfrac{5}{3}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{1}{3}=0\) hay x = \(\dfrac{1}{3}\)

Vậy GTLN của Q bằng \(-\dfrac{5}{3}\) khi x = \(\dfrac{1}{3}\).

b) \(R=2-x^2-y^2-2\left(x+y\right)\)

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

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

\(=-\left(x^2+2x+1+y^2+2y+1-4\right)\)

\(=-\left[\left(x+1\right)^2+\left(y+1\right)^2\right]+4\)

Ta có: \(-\left[\left(x+1\right)^2+\left(y+1\right)^2\right]\le0\forall x,y\Rightarrow-\left[\left(x+1\right)^2+\left(y+1\right)^2\right]+4\le4\forall x,y\)

Dấu "=" xảy ra khi x + 1 = 0 và y + 1 = 0

hay x = -1 và y = -1

Vậy GTLN của R bằng 4 khi x = -1 và y = -1.

c) \(S=-x^2+4x-9\)

\(=-\left(x^2-4x+9\right)\)

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

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

Ta có: \(-\left(x-2\right)^2\le0\forall x\Rightarrow-\left(x-2\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 hay x = 2

Vậy GTLN của S bằng -5 khi x = 2.

Bài 1 :

a) Ta thấy : \(\left(x^2-9\right)^2\ge0\)

                   \(\left|y-2\right|\ge0\)

\(\Leftrightarrow A=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)

Dấu " = " xảy ra :

\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)

Vậy \(Min_A=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)

b) Ta thấy : \(B=x^2+4x-100\)

\(=\left(x+4\right)^2-104\ge-104\)

Dấu " = " xảy ra :

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy \(Min_B=-104\Leftrightarrow x=-4\)

c) Ta thấy : \(C=\frac{4-x}{x-3}\)

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

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

Để C min \(\Leftrightarrow\frac{1}{x-3}\)min

\(\Leftrightarrow x-3\)max

\(\Leftrightarrow x\)max

Vậy để C min \(\Leftrightarrow\)\(x\)max

p/s : riêng câu c mình không tìm được C min :( Mong bạn nào giỏi tìm hộ mình

Bài 2 : 

a) Ta thấy : \(x^2\ge0\)

                  \(\left|y+1\right|\ge0\)

\(\Leftrightarrow3x^2+5\left|y+1\right|-5\ge-5\)

\(\Leftrightarrow C=-3x^2-5\left|y+1\right|+5\le-5\)

Dấu " = " xảy ra :

\(\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy \(Max_A=-5\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)

b) Để B max

\(\Leftrightarrow\left(x+3\right)^2+2\)min

Ta thấy : \(\left(x+3\right)^2\ge0\)

\(\Leftrightarrow\left(x+3\right)^2+2\ge2\)

Dấu " = " xảy ra :

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy \(Max_B=\frac{1}{2}\Leftrightarrow x=-3\)

c) Ta thấy : \(\left(x+1\right)^2\ge0\)

\(\Leftrightarrow x^2+2x+1\ge0\)

\(\Leftrightarrow-x^2-2x-1\le0\)

\(\Leftrightarrow C=-x^2-2x+7\le8\)

Dấu " = " xảy ra :

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy \(Max_C=8\Leftrightarrow x=-1\)

a: \(A=-x^2-4x-2\)

\(=-x^2-4x-4+2\)

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

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

Dấu '=' xảy ra khi x+2=0

=>x=-2

b: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}< =\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{4}=0\)

=>\(x=-\dfrac{3}{4}\)

c: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-x^2-2x-1+9\)

\(=-\left(x^2+2x+1\right)+9\)

\(=-\left(x+1\right)^2+9< =9\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

d: \(D=-8x^2+4xy-y^2+3\)

\(=-8\left(x^2-\dfrac{1}{2}xy\right)-y^2+3\)

\(=-8\left(x^2-2\cdot x\cdot\dfrac{1}{4}y+\dfrac{1}{16}y^2\right)+\dfrac{1}{2}y^2-y^2+3\)

\(=-8\left(x-\dfrac{1}{4}y\right)^2-y^2+3< =3\forall x,y\)

Dấu '=' xảy ra khi y=0 và x-1/4y=0

=>y=0 và x=0

c) 8x³ - 12x^2 + 6x - 1 = 0

⇔ ( 2x - 1 )\(^3\) = 0

⇔ 2x - 1 = 0

⇔ x = \(\frac{1}{2}\)

e) x^3 + 5x^2 + 9x = -45

⇔ x\(^3\) + 5x\(^2\) + 9x + 45 =0

⇔ x\(^2\) ( x + 5 ) + 9( x + 5 ) = 0

⇔ ( x\(^2\) + 9 ) ( x + 5 ) = 0

⇔( x + 3 ) ( x - 3 ) ( x + 5 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)

g) x^2 + 16 = 10x

⇔ x\(^2\) - 10x + 16 = 0

⇔ x\(^2\) - 8x - 2x + 16 = 0

⇔ x( x - 8 ) - 2 ( x - 8 ) = 0

⇔ ( x - 2 ) ( x - 8 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

BÀI 1:

\(A=\left(x-10\right)^2+103\)

Có:    \(\left(x-10\right)^2\ge0\forall x\)

=>   \(A\ge103\)

DẤU "=" XẢY RA <=>   \(\left(x-10\right)^2=0\Rightarrow x=10\)

\(B=\left(2x+1\right)^2-6\)

Có:   \(\left(2x+1\right)^2\ge0\forall x\)

=>   \(B\ge-6\)

DẤU "=" XẢY RA <=>   \(\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

BÀI 3:

a) \(A=y^4+y^3-y^2-2y-\left(y^4+y^3+y^2-2y^2-2y-2\right)\)

\(A=y^4+y^3-y^2-2y-y^4-y^3+y^2+2y+2\)

\(A=2\)

b)   \(B=\left(2x\right)^3+3^3-8x^3+2\)

\(B=29\)

Bài 1.

A = x² - 20x + 103

A = ( x² - 20x + 100 ) + 3

A = ( x - 10 )² + 3 ≥ 3 ∀ x

Đẳng thức xảy ra <=> x - 10 = 0 => x = 10

=> MinA = 3 <=> x = 10

B = 4x² + 4x - 5

B = ( 4x² + 4x + 1 ) - 6

B = ( 2x + 1 )² - 6 ≥ -6 ∀ x

Đẳng thức xảy ra <=> 2x + 1 = 0 => x = -1/2

=> MinB = -6 <=> x = -1/2

Bài 2.

A = -x² + 8x - 21

A = -x² + 8x - 16 - 5

A = -( x² - 8x + 16 ) - 5

A = -( x - 4 )² - 5 ≤ -5 ∀ x

Đẳng thức xảy ra <=> x - 4 = 0 => x = 4

=> MaxA = -5 <=> x = 4

B = lỗi đề :>

Bài 3.

a) y( y³ + y² - y - 2 ) - ( y² - 2 )( y² + y + 1 )

= y⁴ + y³ - y² - 2y - ( y⁴ + y³ + y² - 2y² - 2y - 2 )

= y⁴ + y³ - y² - 2y - y⁴ - y³ - y² + 2y² + 2y + 2

= 2 ( đpcm )

b) ( 2x + 3 )( 4x² - 6x + 9 ) - 2( 4x³ - 1 )

= ( 2x )³ + 27 - 8x³ + 2

= 8x³ + 27 - 8x³ + 2

= 29 ( đpcm )