a) = 2x²y²(3y² - 4y² + 5y)

= 2x²y² * ( - y² +5y)

=2x²y² * y(5-y)

PTĐTTNT

a) 

\(6x^2y^4-8x^2y^2+10x^2y^3\)

\(=x^2y^2\left(6y^2-8+10y\right)\)

b) 

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

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

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

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

c)

\(x^2-25-5+2\sqrt{5}\)

\(=x^2-5^2-5+2\sqrt{5}\)

\(=x^2-5\left(5+1+\sqrt{2}\right)\)

ahihi

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

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

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

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

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

b) \(4x^2+4x+1-y^2-16y-64\)

\(=\left(2x+1\right)^2-\left(y+8\right)^2\)

\(=\left(2x+1-y-8\right)\left(2x+1+y+8\right)\)

\(=\left(2x-7-y\right)\left(2x+9+y\right)\)

c) \(x^3+3x^2+3x+1-27z^3\)

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

\(=\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

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

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

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

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

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

 \(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\)

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

bó tay

Theo bài ra , ta có : 

\(a^8+a+1=a\left(a^7.1\right)=1\)

Sai đề nha tran xuan hoang