Question

1. Phân tích thành nhân tử =phương pháp đặt nhân tử chung a) 4y².(x²+y²) -5x³y-5xy³

b) (x²-x+2).(x-1)-x²(1-x)²-(2x+1).(1-x)³

Answer 1

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

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

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

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

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

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