a.5x²-10xy+5y²-20z²

  =5(x²-2xy+y²-4z²)

  =5[ (x²-2xy+y²)-(2z)² ]

  =5[ (x-y)²-(2z)² ]

  =5(x-y-2z)(x-y+2z)

b.16x-5x²-3

  =15x+x-5x²-3

  =(15x-3)+(x-5x²)

  =3(5x-1)+x(1-5x)

  =3(5x-1)-x(5x-1)

  =(5x-1)(3-x)

c.x²-5x+5y-y²

  =(5y-5x)+(x²-y²)

  =5(y-x)+(x-y)(x+y)

  =5(y-x)-(y-x)(y+x)

  =(y-x)[5-(y+x)]

  =(y-x)(5-y-x)

d.3x²-6xy+3y²-12z²     (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)

=3(x²-2xy+y²-4z²)

=3[ (x²-2xy+y²)-(2z)² ]

=3[ (x-y)²-(2z)² ]

=3(x-y-2z)(x-y+2z)

e.x²+4x+3

=x²+3x+x+3

=(x²+x)+(3x+3)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f.(x²+1)²-4x²

=(x²+1)²-(2x)²

=(x²+1-2x)(x²+1+2x)

h.x²-4x-5

=x²-5x+x-5

=(x²+x)+(-5x-5)

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

-(x+1)(x-5)

5x² - 10xy + 5y² - 20z²

= 5.(x² - 2xy + y² - 4z²)

= 5.[(x² - 2xy + y²) - (2z)²]

= 5.[(x - y)² - (2z)²]

= 5.(x - y - 2z).(x - y + 2z)

x².(1 - x²) - 4 + 4x²

= x².(1 - x²) - 4.(1 - x²)

= (1 - x²).(x² - 4)

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

5x² - 10xy + 5y² - 20z²

= 5.(x² - 2xy + y² - 4z²)

= 5.[(x² - 2xy + y²) - (2z)²]

= 5.[(x - y)² - (2z)²]

= 5.(x - y - 2z).(x - y + 2z)

x².(1 - x²) - 4 + 4x²

= x².(1 - x²) - 4.(1 - x²)

= (1 - x²).(x² - 4)

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

a) phân tích đc \(\left(x+y+2z\right)\left(x+y-2z\right)\)

b) phân tích đc \(\left(1-x^2\right)\left(x-2\right)\left(x+2\right)\)

k chị nha

a/  (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)

b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)

c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)

2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3

b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!

Answer:

Bài 1:

\(5x² - 10xy + 5y² - 20z²\)

\(= 5( x² - 2xy + y² - 4z²)\)

\(= 5 [(x² - 2xy + y²) - (2z)²]\)

\(= 5 [(x - y)² - (2z)²]\)

\(= 5 (x - y - 2z) ( x - y + 2z)\)

\(16x - 5x² - 3 \)

\(= -( 5x² - 16x + 3)\)

\(= -( 5x² - 15x - 1x + 3)\)

\(= - [ (5x² -x) - (15x -3)]\)

\(= - [ x(5x -1) -3(5x -1)]\)

\(= - (5x-1)(x-3)\)

\(x² + 4x + 3\)

\(= x² + x + 3x + 3\)

\(= (x² + x) + (3x + 3)\)

\(= x( x + 1) +3 (x+1)\)

\(= (x+1) (x+3)\)

Bài 2:

\(5x\left(x^2-9\right)=0\)

\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)

Trường hợp 1: \(5x=0\Leftrightarrow x=0\)

Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)

Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)

\(x^2-7x+10=0\)

\(\Rightarrow x^2-5x-2x+10=0\)

\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)

a) \(5x^2+5xy-x-y\)

\(=5x.\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

b) \(5x^2-10y+5y^2-20z^2\)

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

Đề sai ở đâu đó.

c) \(4x^2-y^2+4x+1\)

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

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

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

5x²-10xy+5y²-20z²

=5(x²-2xy+y²-4z²)

=5[(x-y)²-4z²]

=5(x-y-2z)(x-y+2z)

dễ hiểu thôi.hơi vắn tắt 

5x² - 10xy + 5y² - 20z²

=5(x² - 2xy + y² - 4z²)

= 5[ (x² - 2xy + y²) - (2z)²]

= 5[(x-y)² - (2z)²]

= 5(x-y-2z)(x-y+2z)

5x²-10xy+5y²-20z²

=5(x²-2xy+y²-4z²)

=5[(x-y)²-(2z)²]

=5(x-y-2z)(x-y+2z)

a Đề sai: )

b

\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

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

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

c: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

d: =x^2(x^2+2x+1)

=x^2(x+1)^2

e: =5(x^2-2xy+y^2-z^2)

=5[(x-y)^2-z^2]

=5(x-y-z)(x-y+z)

a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)

b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)

c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)

d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)