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) \(x^2-x-y^2-y\)

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

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

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

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

b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)

c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)

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

d)\(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

\(x^2-x-y^2-y\)

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

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

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

hk tốt

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 

a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

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

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

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

1)

a) => 16x² - 8x + 1 - 8(2x² + 3x - 4x - 6) = 15

=> 16x² - 8x + 1 - 8(2x² - x - 6) = 15

=> 16x² - 8x + 1 - 16x² + 8x + 48 = 15

=> 49 = 15 (?) (vô lí)

=> Không tìm được x thoả mãn

b) (5x - 2)(x - 2) - 4(x - 3) = x² + 3

=> 5x² - 10x - 2x + 4 - 4x + 12 = x² + 3

=> 5x² - 16x + 16 = x² + 3

=> 4x² - 16x + 16 = 3

=> (2x)² - 2.2x.4 + 4² = 3

=> (2x - 4)² = 3

=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)

Mong bạn xem lại đề bài!

2) 

a) 5x² - 10xy + 5y² - 20z²

= 5(x² - 2xy + y² - 4z²)

= 5[(x - y)² - (2z)²]

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

b) a³ - ay - a²x + xy

= a(a² - y) - x(a² - y)

= (a - x)(a² - y)

c) 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x - y)² - (2z)²]

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

d) x² - 2xy + tx - 2ty

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

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

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

b: \(5x^2+5xy-x-y\)

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

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

Câu 2 nha

\(a,x^4+2x^3+x^2\)

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

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

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

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

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

\(d,5x^2-10xy+5y^2-20z^2\)

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

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

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

chiều lm tiếp cho