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 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)

\(1.5x^2-10xy+5y^2-20z^2\)

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

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

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

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

\(2.16x-5x^2-3\)

\(=-\left(5x^2-16x+3\right)\)

\(=-\left(5x^2-15x-x+3\right)\)

\(=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]\)

\(=-\left[5x\left(x-3\right)-\left(x-3\right)\right]\)

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

\(3.x^2-5x+5y-y^2\)

\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)

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

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

\(4.3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
\(5.x^2+4x+3\)

\(=x^2+3x+x+3\)

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

\(=x\left(x+3\right)+\left(x+3\right)\)

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

\(6.\left(x^2+1\right)^2-4x^2\)

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

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

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

\(7.x^2-4x-5\)

\(=x^2-5x+x-5\)

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

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

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

a.16x-5x²-3 = - ( 5x²-16x+3) = -( 5x²-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3) 

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

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

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

c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)

d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)

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

e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)

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

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

f.2(x^5)-x^2-5x ( mik ko bik làm)

g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)

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

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

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

a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)

b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y

c,đề bài thiếu  ẩn ở hạng tử thứ nhất ạ !

b mình viết đúng rồi mà, c hạng tử 1 là x^3

câu c bạn dùng máy tính bấm nghiệm nhé , dùng mode =>5=> 4 nhé ! 

còn câu b mình chưa ra !

\(x^2-y^2+5x-5y\)

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

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

\(---\)

\(x^2-16y^2+4x+4\)

\(=\left(x^2+4x+4\right)-16y^2\)

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

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

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

\(---\)

\(3x^2+6xy+3y^2-12\)

\(=3\left(x^2+2xy+y^2-4\right)\)

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

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

\(---\)

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

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

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

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) \(4x^2-y^2+4x+1\)

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

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

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