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

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

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

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) P = 3 - 4x - x²

= -x² - 4x + 3

= -(x² + 4x + 4 - 4) + 3

= -(x + 2)² + 7

Ta có: -(x + 2)² ≤ 0 với ∀x

Nên: -(x + 2)² + 7 ≤ 7 với ∀x

Dấu "=" xảy ra ⇔ -(x + 2)² = 0

x + 2 = 0

x = -2

Vậy GTLN của biểu thức P là 7 khi x = -2

d) S = -x² + 4x - 9

= -(x² - 4x + 4 - 4) - 9

-(x - 2)² - 5

Ta có: -(x - 2)² ≤ 0 với ∀x

Nên: -(x - 2)² - 5 ≤ -5 với ∀x

Dấu "=" xảy ra ⇔ -(x - 2)² = 0

x - 2 = 0

x = 2

Vậy GTLN của biểu thức S là -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

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

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

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

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

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

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

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

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

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

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

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

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

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

\(A\left(x\right)=-\left(x^2-\frac{5}{3}x\right)+1=-3\left(x^2-2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2\right)+1+3.\left(\frac{5}{6}\right)^2\)

\(=-3\left(x-\frac{5}{6}\right)^2+\frac{37}{12}\le\frac{37}{12}\)

Dấu "=" xảy ra khi \(x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy GTLN của A là 37/12.

b, c làm tương tự.