B A C D E K I I' D'

Từ I vẽ đường thẳng II' // BC

Từ D vẽ đường thẳng DD' // BC

=> II' // DD' . Mà I là trung điểm của DE

=> EI' = I'D' ( 1 )

Vì \(\Delta\)ABC cân tại A có DD' // BC => DB = D'C ( 2 )

Mà AD = CE => AE = DB ( 3 )

Từ ( 2 ) và ( 3 ) => D'C = AE ( 4 )

Từ ( 1 ) và ( 4 ) => AI' = 'IC

\(\Delta\)AKC có II' // KC ; AI' = I'C

=>AI = IK ( Đpcm )

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 

\(4x-4y+x^2-2xy+y^2\)   

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

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