Câu 1:

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

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

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

=x²(x-3)-4x+3.4

=x²(x-3)-4(x+3)

=x²(x-3)+4(x-3)

=(x-3)(x²+4)

=(x-3)(x²+2²)

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

Viết liền thế ai mà đọc được

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

\(x^3-4x^2+8x-8\)

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

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

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

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

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

Câu này số xấu không phân tích được nhân tử lun

\(x^3\left(2+x\right)^2-\left(x+2\right)^2+1-x^3\\ =\left(x+2\right)^2\left(x^3-1\right)-\left(x^3-1\right)\\ =\left[\left(x+2\right)^2-1\right]\left(x^3-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x^2+x+1\right)\)

bạn có thể viết rõ hơn ko

x^4+x^2y^2+y^4

=x^4+2x^2y^2+y^4-x^2y^2

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

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

= ( x² - xy) + (x - y)

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

= (x+1)(x-y)

x² - 4y² + x + 2y

= ( x² - 4y² ) + ( x + 2y )

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

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