A = (3x-2)^2-(x+3)^2

= 9x^2 - 12x + 4 - x^2 - 6x - 9

= 8x^2 - 18x - 5

B = (5x+3)^2+(x-2)^2

= 25x^2 + 30x + 9 + x^2 - 4x + 4

= 26x^2 +26x +13

C = (2x+y-3)^2-(x+2y+3)^2

= (2x + y)^2 - 6(2x + y) + 9 - (x + 2y)^2 - 6(x + 2y) - 9

= 4x^2 + 4xy + y^2 - 12x - 6y - x^2 - 4xy - 4y^2 - 6x - 12y 

= 3x^2 - 3y^2 -18x - 18y

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

= (x + 2y)^2 + 6z(x + 2y) + 9z^2 - (x - 2y)^2 + 6z(x - 2y) - 9z^2

= x^2 + 4xy + y^2 + 6xz + 12yz - x^2 + 4xy - y^2 + 6xz - 12yz 

= 8xy + 12xz 

A= (3x-2)^2-(x+3)^2

= (9x^2 - 12x + 4) - (x^2 + 6x +9)

= 8x^2 - 6x + 13

Chết rùi! Quên đổi dấu => làm sai/

a) 3x(x + 1) - 5y(x + 1)

= (x + 1)(3x - 5y)

b) 3x(x - 6) - 2(x - 6)

= (x - 6)(3x - 2)

c) 4y(x - 1) - (1 - x)

= 4y(x - 1) + (x - 1)

= (x - 1)(4y + 1)

d) (x - 3)³ + 3 - x

= (x - 3)³ - (x - 3)

= (x - 3)[(x - 3)² - 1]

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

= (x - 3)(x - 4)(x - 2)

e) 7x(x - y) - (y - x)

= 7x(x - y) + (x - y)

= (x - y)(7x + 1)

h) 3x³(2y - 3z) - 15x(2y - 3z)²

= (2y - 3z)[3x³ - 15x(2y - 3x)]

= 3x(2y - 3x)[x² - 5(2y - 3x)]

= 3x(2y - 3x)(x² - 10y + 3x)

= 3x(2y - 3x)(x² + 3x - 10y)

k) 3x(x + 2) + 5(-x - 2)

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

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

l) 18x²(3 + x) + 3(x + 3)

= (x + 3)(18x² + 3)

= 3(x + 3)(6x² + 1)

m) 7x(x - y) - (y - x)

= 7x(x - y) + (x - y)

= (x - y)(7x + 1)

n) 10x(x - y) - 8y(y - x)

= 10x(x - y) + 8y(x - y)

= (x - y)(10x + 8y)

= 2(x - y)(5x + 4y)

h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)

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

l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)

m) \(=7xy\left(2x-3y+4xy\right)\)

n) \(=2\left(x-y\right)\left(5x-4y\right)\)

Ý a có rì đó sai sai nha bn 

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

Bài 2: 

\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

⇔⎡⎣x=1x=−13

e) Ta có: \(a^3-a^2-a+1\)

\(=a^2\left(a-1\right)-\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2-1\right)\)

\(=\left(a-1\right)^2\cdot\left(a+1\right)\)

f) Ta có: \(x^3-2xy-x^2y+2y^2\)

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

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

a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)

b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

d) Đề sai ko ???

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

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

a, \(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(\left(a-b\right)\left(a+b\right)\right)^2=\left(a^2-b^2\right)^2\)

\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

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

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

\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)

\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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