a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy Max(A) = 7 khi x = -1

b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy Max(B) = 11 khi x = 3

c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{1}{3}\right)^2=0\Rightarrow x=\frac{1}{3}\)

Vậy Max(C) = 1/3 khi x = 1/3

d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy Max(D) = 17/4 khi x = 3/2

e) \(E=3-2x^2+2xy-y^2-2x\)

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

\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)

Vậy Max(E) = 4 khi x = y = -1

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

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

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

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

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

\(=2x\cdot2x=4x^2\)

b) \(2x^2-2xy-4y^2\)

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

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

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

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

c) \(x^2-2x-4y^2-4y\)

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

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

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

d) \(4x\left(x-2y\right)-8y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(4x-8y\right)\)

\(=4\left(x-2y\right)\left(x-2y\right)\)

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

x² + 2y² + z² - 2xy - 2y - 4z + 5 = 0

<=> ( x² - 2xy + y² ) + ( y² - 2y + 1 ) + ( z² - 4z + 4 ) = 0

<=> ( x - y )² + ( y - 1 )² + ( z - 2 )² = 0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )² + ( y - 1 )² + ( z - 2 )²\(\ge\)0\(\forall\)x ; y ; z

Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )

Thay ( 1 ) vào A , ta được :

\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)

Vậy A = 2

Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)

Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)

Ta có: \(x^2+x-6=\left(x-2\right)\left(x+3\right)\)

Đặt \(A\left(x\right)=x^3+ax^2-bx+12\) 

Để A(x) chia hết cho \(x^2+x-6\) thì mọi nghiệm của \(x^2+x-6\) đều là nghiệm của A(x)

=> x = 2 và x = -3 là 2 nghiệm của A(x)

Ta có: \(\hept{\begin{cases}A\left(2\right)=2^3+4a-2b+12=0\\A\left(-3\right)=\left(-3\right)^3+\left(-3\right)^2a-\left(-3\right)b+12=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}4a-2b=-20\\9a+3b=15\end{cases}}\Leftrightarrow\hept{\begin{cases}2a-b=-10\\3a+b=5\end{cases}}\)

\(\Rightarrow2a-b+3a+b=-10+5\)

\(\Leftrightarrow5a=-5\Rightarrow a=-1\Rightarrow b=8\)

Vậy a = -1 ; b = 8

1) \(x^5-x^4-1\)

\(=x^5-x^4-1+x^3-x^3+x^2-x^2+x-x\)

\(=\left(x^5-x^3-x^2\right)-\left(x^4-x^2-x\right)+\left(x^3-x-1\right)\)

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

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

2) \(x^8-3x^4+1\)

\(=x^8-3x^4+1+x^4-x^4\)

\(=\left(x^8-2x^4+1\right)-x^4\)

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

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

a. x⁴ - 27x = x ( x³ - 3³ ) = = x ( x - 3 ) ( x² + 3x + 3² ) = x ( x - 3 ) ( x² + 3x + 9 )

b. x³ + 2x² + 2x + 1 = ( x³ + 1³ ) + ( 2x² + 2x ) = ( x + 1 ) ( x² - x + 1 ) + 2x ( x + 1 ) = ( x + 1 ) ( x² + x + 1 )

c. 4x - 4y + x² - 2xy + y² = 4 ( x - y ) + ( x - y )² = ( x - y ) ( x - y + 4 )

a 

\(x^4-27x\)   

\(=x\left(x^3-27\right)\)   

\(=x\left(x^3-3^3\right)\)   

\(=x\left(x-3\right)\left(x^2+3x+9\right)\)   

b 

\(x^3+2x^2+2x+1\)   

\(=x^3+x^2+x^2+x+x+1\)   

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

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

c 

\(4x-4y+x^2-2xy+y^2\)   

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

\(=\left(x-y\right)\left(x-y+4\right)\)

A = -5 - (x - 1)(x + 2)

   = -5 - [x(x + 2) - 1(x + 2)]

   = -5 - (x² + 2x - x - 2)

   = -5 - x² - 2x - x + 2 = -5 - x² - x + 2 = (-5 + 2) - x² - x = -3 - x² - x

  = -(x + x² + 3) = -(x² + x + 3)

  = -[x² + 2.x.1/2 + (1/2)² ] - 11/4

  = -(x + 1/2)² - 11/4

Vì (x + 1/2)² \(\ge\)0\(\forall\)x

=> -(x + 1/2)² \(\le\)0\(\forall\)x

=> \(-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}\ge-\frac{11}{4}\forall x\)

Dấu " = " xảy ra khi (x + 1/2)² = 0 => x = -1/2

Vậy A_max = -11/4 khi x = -1/2

\(A=-5-\left(x-1\right)\left(x+2\right)\)   

\(=-5-\left(x^2+x-2\right)\)   

\(=-5-x^2-x+2\)   

\(=-x^2-x-3\)   

\(=-x^2-x-\frac{1}{4}-3+\frac{1}{4}\)   

\(=-\left(x^2+x+\frac{1}{4}\right)-\frac{11}{4}\)   

\(=-\left(x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right)-\frac{11}{4}\)    

\(=-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}\)   

Ta có \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)   

\\(-\left(x+\frac{1}{2}\right)^2\le0\)   

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

Dấu = xảy ra 

\(\Leftrightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

A = ( x + y )( x + 2y )( x + 3y )( x + 4y ) + y⁴

= [ ( x + y )( x + 4y ) ][ ( x + 2y )( x + 3y ) ] + y⁴

= ( x² + 5xy + 4y² )( x² + 5xy + 6y² ) + y⁴ (1)

Đặt t = x² + 5xy + 5y²

(1) <=> ( t - y² )( t + y² ) + y⁴

       = t² - y⁴ + y⁴

       = t² = ( x² + 5xy + 5y² )²

Vì x, y nguyên => x² nguyên ; 5xy nguyên ; 5y² nguyên

=> x² + 5xy + 5y² nguyên

=> ( x² + 5xy + 5y² )² là một số chính phương

=> đpcm

A = ( x + y )( x + 2y )( x + 3y )( x + 4y ) + y⁴ 

=> A = ( x² + 5xy + 4y² ) ( x² + 5xy + 6y² ) + y⁴

Đặt a = x² + 5xy + 5y² , pt trở thành :

A = ( a - y² ) ( a + y² ) + y⁴

=> A = t² - y⁴ + y⁴ = t² = ( x² + 5xy + 5y² )² là SCP

Vậy A là SCP

ê ko ý là t thấy trên kia có ng lm r á :>