Question

Phân tích đa thức thành nhân tử:

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

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

3) C = \(\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2\)

Answer 1

\(1,\left(x+2y-3\right)^2-4\left(x+2y-3\right)+4=\left(x+2y-3-2\right)^2=\left(x+2y-5\right)^2\)

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

\(3,\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2=\left(x^2+y^2-17\right)-\left(2xy-8\right)^2=\left(x^2-2xy+y^2-9\right)\left(x^2+y^2+2xy-25\right)=\text{[}\left(x-y\right)^2-3^2\text{]}\text{[}\left(x+y\right)^2-5^2\text{]}=\left(x-y+3\right)\left(x-y-3\right)\left(x+y+5\right)\left(x+y-5\right)\)