a)

\(2x^2y-8xy^2\\ =2xy\left(x-4y\right)\)

b)

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

c: \(x^2-4+3\left(x-2\right)^2\)

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

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

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

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

a) Ta có: \(x^2-2xy+y^2-2x+2y\)

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

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

b) Ta có: \(x^2-4x+4-x^2y+2xy\)

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

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

c) Ta có: \(ax^2-3axy-x^2+6xy-9y^2\)

\(=ax\left(x-3y\right)-\left(x^2-6xy+9y^2\right)\)

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

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

d) Ta có: \(2a^2x-5a^2y-4x^2+30xy-25y^2\)

\(=a^2\left(2x-5y\right)-\left(4x^2-30xy+25y^2\right)\)

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

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

1, x²(x²+2x+1)=x²(x+1)²

2, 2(x²+2x+1-y²)=2(x+1-y)(x+1+y)

3, 16-(x²+2xy+y²)=(4-x-y)(4+x+y)

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

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

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

hk tốt

^^

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

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

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

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

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

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

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

Bài giải:

a) x³ – 2x² + x = x(x² – 2x + 1) = x(x – 1)²

b) 2x² + 4x + 2 – 2y² = 2[(x² + 2x + 1) – y²]

= 2[(x + 1)² – y²]

= 2(x + 1 – y)(x + 1 + y)

c) 2xy – x² – y² + 16 = 16 – (x² – 2xy + y²) = 4² – (x – y)²

= (4 – x + y)(4 + x – y)

a) \(x^3 - 2x^2 + x\) \(= x(x^2 - 2x + 1)\)

\(= x (x - 1 )^2\)

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

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

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

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

c) \(2xy - x^2 - y^2 + 16\) \(= - (x^2 - 2xy + y^2 - 4^2)\)

\(= - [(x^2 - 2xy + y^2) - 4^2]\)

\(= - [(x-y)^2 - 4^2 ]\)

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